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This question appeared in week four relays at mathcamp. I solved it and really liked it so I'm giving it to you to try to figure out. The rules are no calculators, computers, or turing machines.

if $a_{0}= 8^{7^{6^{5^{4^{3^{2^{1^{0}}}}}}}}$ and $a_{n+1}$ is the sum of the digits of $a_{n}$, then what is $a_{100}$?

Is the answer 8??I hav interpreted the sum of the digits to be the digit sum which is infact number modulo 9...

I'm getting that

$7^{6^{5^{4^{3^{2^{1^{0}}}}}}}$ is odd, and so

$8^{7^{6^{5^{4^{3^{2^{1^{0}}}}}}}}$ is congruent to

$(-1)^{k}$ mod 9 where $k$ is odd and so its congruent to 8...

but how do we prove that it is actually 8 and not just congruent to 8?

The answer is indeed [hide]8[/hide]

To prove it is what it is, think about how the number changes from one step to another.

[quote="Varun Nakra"]The Digit sum is ofcourse 8 only..But if we are to calculate the actual sum of the digits which is not infact the same as the digit sum..Then is there a way out :maybe:[/quote]

The sum of the digits isn't the same as the digit sum? :huh:

[quote="E^(pi*i)=-1"]Hello, Hamster1800!

Ah, yes, I remember this problem.  I think my team got it right, but I don't remember what answer we got then.

[hide]

We use some very weak upper bounds.  We know that

$3^{2^{1^{0}}}=9$, so we can write the expression as

$8^{7^{6^{5^{4^{9}}}}}$

Each of the digits is less than 10, so we can rewrite the expression as

$10^{10^{10^{10^{10^{10}}}}}$

We know that the original expression is less than this. The largest digit sum that a number less than this can have is

$9(10^{10^{10^{10^{10}}}})$

We can just regard the 9 as a 10 and thus add 1 to the exponent.  The largest possible digit sum then is

$9(10^{10^{10^{10}}}+1)$

Similarly, the largest possible digit sum of this is

$9(10^{10^{10}}+1)$

Two digit sums later we get $9(10+1)=99$.  From here we get 18, and then 9.  This stays the same until $a_{100}$, so we know that the digit sum of the original expression is $\leq 9$.

Now we can use tim1234133's suggestion to see that the digit sum is in fact 8.

[/hide][/quote]

Oooh!! This is who you are!!

Anyway, I'd like to explain my solution below:

[hide]
First, we know that $9 = 3^{2}$ so $\varphi(9) = (9-3) = 6$. If you don't know what the Euler phi function is, $\varphi(n)$ is the number of numbers less than or equal to $n$ that are relatively prime to $n$. This function satisfies the nice property that $a^{\varphi(n)}\equiv 1 \pmod n$ provided that $\gcd(a,n) = 1$. Since $\gcd(8,9) = 1$, we can apply this property and reduce $7^{6^{5^{4^{3^{2^{1^{0}}}}}}}$ mod 6.

EDIT: oops, I just realized that I made a mistake and it should be that $\varphi(9) = 6$, not 4.

Now, here is where I just saw a nicer solution than what I had before. $\varphi(6) = \varphi(2\cdot3)= (3-1)(2-1) = 2$ So, we just reduce $6^{5^{4^{3^{2^{1^{0}}}}}}$ mod 2 (and we can because $\gcd(7,6)=1$) , which is $0$ since $6\equiv 0 \pmod 2$ and $0^{k}= 0$. So, we now have that $8^{7^{6^{5^{4^{3^{2^{1^{0}}}}}}}}\equiv 8^{7^{0}}\equiv 8 \pmod 9$.

Now we have to think whether it is actually 8 or some multiple of 9 greater than 8. Every time you sum the digits of a number, the number gets reduced to a number that depends on the [i] length[/i] of the input, meaning that the output is about the size of $\log n$. Since $8^{7^{6^{5^{4^{3^{2^{1^{0}}}}}}}}$ is much less than a power tower of 100 10s, it will reduce to one digit. So the answer is $\boxed{8}$.

[/hide]

Also, according to mathword, the digit sum is the sum of the digits, while the digital [b]root[/b] is the sum of the digits of the sum of the digits of the sum of the digits of the sum of the digits of the ... of the sum of the digits where you repeat taking the digit sum until you are left with one digit.

EDIT: Oh yeah I forgot something:

[hide="only look if you have read the solution"]
[hide="how to calculate the phi function"]
I told you what the phi function was defined as, but I forgot to mention the technique I used to calculate it.

If you know the factorization of a number, $n = p_{1}^{k_{1}}p_{2}^{k_{2}}p_{3}^{k_{3}}\cdots p_{r}^{k_{r}}$ or in product notation $n = \prod_{i=1}^{r}p_{i}^{k_{i}}$ where all the $p_{i}$'s are prime, then the Euler Phi Function is calculated as: $\varphi(n) = (p_{1}^{k_{1}}-p_{1}^{k_{1}-1})(p_{2}^{k_{2}}-p_{2}^{k_{2}-1})(p_{3}^{k_{3}}-p_{3}^{k_{3}-1})\cdots(p_{r}^{k_{r}}-p_{r}^{k_{r}-1})$.

You can also take a $p_{i}^{k_{i}}$ out of each of the factors and get that $\varphi(n) = n(1-\frac{1}{p_{1}})(1-\frac1{p_{2}})\cdots(1-\frac1{p_{r}})$.
[/hide]
[/hide]

[quote="Varun Nakra"][quote="Hamster1800"][quote="E^(pi*i)=-1"]Hello, Hamster1800!

Ah, yes, I remember this problem.  I think my team got it right, but I don't remember what answer we got then.

[hide]

We use some very weak upper bounds.  We know that

$3^{2^{1^{0}}}=9$, so we can write the expression as

$8^{7^{6^{5^{4^{9}}}}}$

Each of the digits is less than 10, so we can rewrite the expression as

$10^{10^{10^{10^{10^{10}}}}}$

We know that the original expression is less than this. The largest digit sum that a number less than this can have is

$9(10^{10^{10^{10^{10}}}})$

We can just regard the 9 as a 10 and thus add 1 to the exponent.  The largest possible digit sum then is

$9(10^{10^{10^{10}}}+1)$

Similarly, the largest possible digit sum of this is

$9(10^{10^{10}}+1)$

Two digit sums later we get $9(10+1)=99$.  From here we get 18, and then 9.  This stays the same until $a_{100}$, so we know that the digit sum of the original expression is $\leq 9$.

Now we can use tim1234133's suggestion to see that the digit sum is in fact 8.

[/hide][/quote]

Oooh!! This is who you are!!

Anyway, I'd like to explain my solution below:

[hide]
First, we know that $9 = 3^{2}$ so $\varphi(9) = 2(3-1) = 4$. If you don't know what the Euler phi function is, $\varphi(n)$ is the number of numbers less than or equal to $n$ that are relatively prime to $n$. This function satisfies the nice property that $a^{\varphi(n)}\equiv 1 \pmod n$ provided that $\gcd(a,n) = 1$. Since $\gcd(8,9) = 1$, we can apply this property and reduce $7^{6^{5^{4^{3^{2^{1^{0}}}}}}}$ mod 4.

Now, here is where I just saw a nicer solution than what I had before. $\varphi(4) = \varphi(2^{2}) = 2(2-1) = 2$ So, we just reduce $6^{5^{4^{3^{2^{1^{0}}}}}}$ mod 2 (and we can because $\gcd(7,4)=1$) , which is $0$ since $6\equiv 0 \pmod 2$ and $0^{k}= 0$. So, we now have that $8^{7^{6^{5^{4^{3^{2^{1^{0}}}}}}}}\equiv 8^{7^{0}}\equiv 8 \pmod 9$.

Now we have to think whether it is actually 8 or some multiple of 9 greater than 8. Every time you sum the digits of a number, the number gets reduced to a number that depends on the [i] length[/i] of the input, meaning that the output is about the size of $\log n$. Since $8^{7^{6^{5^{4^{3^{2^{1^{0}}}}}}}}$ is much less than a power tower of 100 10s, it will reduce to one digit. So the answer is $\boxed{8}$.

[/hide]

[b]Also, according to mathword, the digit sum is the sum of the digits, while the digital [b]root[/b] is the sum of the digits of the sum of the digits of the sum of the digits of the sum of the digits of the ... of the sum of the digits where you repeat taking the digit sum until you are left with one digit[/b].[/quote]
Yes thats wat i mean..If the quesn is to find the digital sum which u think it is the repeated sum of the digits uptil we get a single digit..then undoubtedly it is 8..Wat i was saying was that..wat if we dnt do that repeatedly..Addin all the digits of the number..Ofcourse ,since the problem was to calculate a100..uptil which,the sum does get reduced to a single digit only.But wat if want the sum of the digits of the number a0 itself..Then??????[/quote]

Then we'd've not been asked to do it with only pencil and paper.

[quote="Hamster1800"][/quote][quote="Varun Nakra"][quote="Hamster1800"][quote="E^(pi*i)=-1"]Hello, Hamster1800!

[/quote]

[b]Also, according to mathword, the digit sum is the sum of the digits, while the digital [b]root[/b] is the sum of the digits of the sum of the digits of the sum of the digits of the sum of the digits of the ... of the sum of the digits where you repeat taking the digit sum until you are left with one digit[/b].[/quote]
Yes thats wat i mean..If the quesn is to find the digital sum which u think it is the repeated sum of the digits uptil we get a single digit..then undoubtedly it is 8..Wat i was saying was that..wat if we dnt do that repeatedly..Addin all the digits of the number..Ofcourse ,since the problem was to calculate a100..uptil which,the sum does get reduced to a single digit only.But wat if want the sum of the digits of the number a0 itself..Then??????[/quote][quote="Hamster1800"]

Then we'd've not been asked to do it with only pencil and paper.[/quote]
[b]Then find the last two digits of the number a0.with only pencil and paper :D [/b]

[quote="Varun Nakra"][/quote][quote="Hamster1800"][/quote][quote="Varun Nakra"][quote="Hamster1800"][quote="E^(pi*i)=-1"]Hello, Hamster1800!

[/quote]

[b]Also, according to mathword, the digit sum is the sum of the digits, while the digital [b]root[/b] is the sum of the digits of the sum of the digits of the sum of the digits of the sum of the digits of the ... of the sum of the digits where you repeat taking the digit sum until you are left with one digit[/b].[/quote]
Yes thats wat i mean..If the quesn is to find the digital sum which u think it is the repeated sum of the digits uptil we get a single digit..then undoubtedly it is 8..Wat i was saying was that..wat if we dnt do that repeatedly..Addin all the digits of the number..Ofcourse ,since the problem was to calculate a100..uptil which,the sum does get reduced to a single digit only.But wat if want the sum of the digits of the number a0 itself..Then??????[/quote][quote="Hamster1800"]

Then we'd've not been asked to do it with only pencil and paper.[/quote][quote="Varun Nakra"]
[b]Then find the last two digits of the number a0.with only pencil and paper :D [/b][/quote]

Ok let's see...We're working mod 100, so since gcd(100,8) = 4, we can't directly apply the phi function. However.....We know that since 8 is a multiple of four, the power tower is a multiple of 4. And since 100 is a multiple of four, what it's congruent to is a multiple of four since we're subtracting a multiple of four from a multiple of four. Therefore, we write:

$a_{0}\equiv x \equiv 4y \pmod{100}$
$a_{0}\equiv 4y \pmod{100}$
$8^{7^{6^{5^{4^{3^{2^{1^{0}}}}}}}-1}\cdot2\cdot4 \equiv 4y \pmod{100}$
$8^{7^{6^{5^{4^{3^{2^{1^{0}}}}}}}-1}\cdot2 \equiv y \pmod{25}$

Now we know that $\varphi(25) = (25-5) = 20$, so we reduce the exponent mod 20.

We also know that $\varphi(20) = \varphi(2^{2}\cdot5) = (4-2)(5-1) = 8$, so we can reduce the exponent on the 7 mod 8.

We know that $6^{5^{4^{3^{2^{1^{0}}}}}}\equiv 0 \pmod 8$ because $5^{4^{3^{2^{1^{0}}}}}> 3$, so there are at least 3 factors of two in the power tower.

So, we have $y \equiv 8^{7^{0}-1}\cdot2 \pmod{25}$

-----From here I made a mistake-----

so $y \equiv 8^{6}\cdot 2 \pmod{25}$

We can expand this by hand:

$8\cdot 8 = 64 \equiv 14 \pmod{25}$
$14\cdot 8 = 112 \equiv 12 \pmod{25}$
$12\cdot 8 = 96 \equiv 21 \equiv-4 \pmod{25}$
$-4\cdot 8 =-32 \equiv-7 \pmod{25}$
$-7 \cdot 8 =-56 \equiv-6 \equiv 8^{6}\equiv 19 \pmod{25}$

$y \equiv 19 \cdot 2 = 38 \equiv 13 \pmod{25}$.

Since we defined $a_{0}= x = 4y$, we know that $x \equiv 13\cdot 4 \equiv 52 \equiv 2 \pmod{25}$. Therefore, the last two digits are either 02, 27, 52, or 77. Since we know that $x$ is a multiple of 4, we know that the last two digits also have to be a multiple of 4. Therefore, the last two digits are $\boxed{52}$.

----- Here is what it should have been -----

so $y \equiv 8^{0}\cdot 2 \equiv 2 \pmod{25}$

Since we defined $x = 4y$, $x \equiv 8 \pmod{25}$. Therefore, the last two digits of x are either 08, 33, 58, or 83. The only one of these divisible by 4 is $\boxed{08}$

-----End Revision-----
Do you guys agree?

[b]Then find the last two digits of the number a0.with only pencil and paper :D [/b][/quote]
[b]Do you guys agree?[/[/b]quote]
Nopes..I dnt agree :D ..The answer should be 8..I mean 08....

[quote="Varun Nakra"]
[quote="Hamster1800"]
[quote="Varun Nakra"][b]Then find the last two digits of the number a0.with only pencil and paper :D [/b][/quote]
Do you guys agree?[/quote]
Nopes..I dnt agree :D ..The answer should be 8..I mean 08....[/quote]
Reasoning for the answer being 08?

Oh doi.. I messed up. I said that $7^{0}= 7$ instead of $7^{0}= 1$. You are right. If you continue my solution with correct arithmetic you do indeed get the answer as 08.

[quote="Hamster1800"][/quote][quote="Varun Nakra"]
[quote="Hamster1800"]
[quote="Varun Nakra"][b]Then find the last two digits of the number a0.with only pencil and paper :D [/b][/quote]
Do you guys agree?[/quote]
Nopes..I dnt agree :D ..The answer should be 8..I mean 08....[/quote][quote="Hamster1800"]
Reasoning for the answer being 08?

Oh doi.. I messed up. I said that $7^{0}= 7$ instead of $7^{0}= 1$. You are right. If you continue my solution with correct arithmetic you do indeed get the answer as 08.[/quote]

Look..U can write x=8^7^6^5^4^3^2^1 as
x=[8^(7)]^{7^(6^5^4^3^2^1-1)}..
Now if one squares 8 we get the last two digits as 64..so 8^7 gives the last two digits as (64)(64)(64)(08) which is infact 52..

so now x=[52^(7)]^{7^(6^5^4^3^2^1-2)}..
Square 52..then u find that the last two digits of 52^7 will be 28

so now x=[28^(7)]^{7^(6^5^4^3^2^1-3)}..

Again repeat the above operation..then last two digits will be 12.
Again doing the same thing we can find that the last two digits will be 08....thus finally

x=[8^(7)]^{7^(6^5^4^3^2^1-4)}..
Thus we can conclude that the last two digits are following a cyclicity of 4..viz.
08=>52=>28=>12=>08 nd so on...
Thus whenver we will have the power mod 4=0..Then the last two digits will be 08..
The power that we have is (6^5^4^3^2^1) mod 4=0..
Hence the last two digits should be 08

It may be simpler than you think..Once you realize $a_{5}$ is one digit number number,   (less than 10) and it is   {8 mod 9} , then $a_{6}=a_{7}=a_{1}00=a_{7}67453453 = 8$

[quote="Hamster1800"]I like both of your solutions. However, @Varun: How long does it take you to find the last 2 digits of 8^7, 52^7, 28^7, AND 12^7 by hand?

@matt. What you were doing when you wanted to show that if $a\equiv b \pmod{20}$ then $2^{a}\equiv 2^{b}\pmod{25}$ can be done by the Euler phi function since $\varphi(25) = 20$.[/quote]
Not more than 2 to 3 mins buddy..You just need to square the numbers and that too not fully and then multiply their last two digits..I used a faster way of squaring them using Vedic Mathematics :D which I guess must be known to you all..

[quote="Gyan"]It may be simpler than you think..Once you realize $a_{5}$ is one digit number number,   (less than 10) and it is   {8 mod 9} , then $a_{6}=a_{7}=a_{100}=a_{767453453}= 8$[/quote]

Well, isn't it a lot harder to realise $a_{5}$ is a one digit number than it is to realise that $a_{100}$ is a one digit number? Since this was a fast paced competition (as many points as possible in an hour), the problems were made to be easily computable by hand.

Also, what's Vedic Mathematics?

EDIT: I did a google search on it and it seems interesting, but Art Benjamin also gave a talk at MathCamp on how to do mental math fast. Let me see if I can remember his method for squaring:

I think he did 52 = 50+2, so 52^2 = 50^2 + 2^2 + 2*2*50 = 2500 + 4 + 200 = 2704

Or you could also do:

52*52. 52*52 = 50*54 + 2^2 = 2700 + 4 = 2704.

He was quite impressive. He even managed to square a 5 digit number mentally!

[quote]Well, isn't it a lot harder to realise a_{5} is a one digit number than it is to realise that a_{100} is a one digit number? Since this was a fast paced competition (as many points as possible in an hour), the problems were made to be easily computable by hand.
[/quote]
Of course, it may be far from easy for $a_{5}$  but it is easy (almost trivial) fo a little larger number, say  for $a_{8}$ or for luck give extra one or two and take for  $a_{10}$ .. [b]It has to be a single digi[/b]t .. and all the rest after that have to be single digit...
and the single digit has to be 8 (becuase thats the only number less than 10 which is 8 modulo 9 (remember  8 = -1 mod 9 and the power (7 raise to something) is odd.. odd power of -1 is always -1)

(All this have already been discussed in previous posts .. so nothing new here in this post_

One chalanging question here is to find out what are the last three digits of this expression.... I will  wait for a day or so before putting answer here.

Finding the last $n$ digits of some large power tower isn't so hard; you apply CRT with $2^{n}, 5^{n}$ and repeatedly reduce each part of the tower $\bmod \varphi(2^{n}), \bmod \varphi( \varphi(2^{n}))$, etc.

As to the original problem, it's painfully easy.  As tim noted, $8^{k}$ where $k$ is odd is congruent to $8 \bmod 9$, so that is the [b]digital root[/b] (easy lemma.)  I don't know why you even needed to introduce the $\varphi$ function to solve the original problem.

Let $s(n)$ denote the sum of the digits of $n$ and let $d(n)$ denote the number of digits of $n$.  It's easy to see that

$d(n) = \lfloor \log_{10}n+1 \rfloor$

And also that

$s(n) \le 9 \cdot d(n)$

It's a simple calculation (use some very weak bounds!) from there to show that, for example, $a_{10}< 10$ from which it follows $a_{10}, a_{11}, ... a_{100}= 8$.

Edit:  Aah, I did not notice that my concerns had already been addressed so early in the thread!  (Why the need for all the other posts?)  The fact that Hamster1800 managed to be oblivious both to the fact that $8^{2m+1}\equiv 8 \bmod 9$ and that $7^{k}\equiv 1 \bmod 6$ was frankly worrying and I felt it needed to be addressed  :oops:

haha. I did the exponents in the wrong order. But I got it now.

 :blush:

[quote="t0rajir0u"]
Edit:  Aah, I did not notice that my concerns had already been addressed so early in the thread!  (Why the need for all the other posts?)  The fact that Hamster1800 managed to be oblivious both to the fact that $8^{2m+1}\equiv 8 \bmod 9$ and that $7^{k}\equiv 1 \bmod 6$ was frankly worrying and I felt it needed to be addressed  :oops:[/quote]
I was posting the solution I used during the one hour "race" and during that time I was reducing all the powers mod 9 instead of $\varphi(9)$,$\varphi(\varphi(9))$, $\varphi(\varphi(\varphi(9)))$, etc, so since I thought I was reducing $7^{6^{5^{4^{3^{2^{1^{0}}}}}}}$ mod 9, I thought it could change parity, so I went further. In fact, I got the wrong answer the first time I attempted this problem because I was reducing that mod 9 and it -did- change parity.

Also, finding the last 2-3 digits of a power tower isn't hard, but how hard is finding the last 4? 5? 6? 7? 8? 9? 10? How fast does it become harder?

Vedic Maths..as the name suggets is composed of "vedas"..vedas were the sutras or the short couplets used to calculate fast.The language was Sanskrit..It had been there in India for aeons..It can be used efficiently anywhere once mastered..Like we need to find the square of (128)..Now I say one thing..If we have a variable D..then D(a)=Sqaure(a)
D(ab)=2ab..D(abc)=b^2+2ac...D(abcd)=2bc+2ad..Now we write
(128)^2=(D(1)/D(12)/D(128)/D(28)/D(8))...Where we need to caluclate the D's of these numbers nd the carry over is added always.Here D(8)=64.hence we wud put down the last digit as 4,then 6 will be the carry over..Then D(28)=2*2*8..so 32+6=38.whose last digit is put down as 8 nd 3 is the carry over..then D(128)=2^2+(2)(1)*(8)=20.so 20+3=23..whose last digit is put down as 6 nd the carry over is 2..Similarly try finding the Ds of each of them nd combine all the last digits to get the answer is 16384..All this can be done orally to get the answer in minutes.. :)

***post deleted by author no longer needed *next message can be deleted too...

[quote="Gyan"]Varun - You are quoting me above, I  don't recall posting those words.. care to edit out your post.  thanks.[/quote]
I have edited sir!! Sorry for the same
 :)
Regards
varun


What is the simplified numerical value of $ \frac{a\plus{}11b}{a\minus{}b}$ if $ \frac{4a\plus{}3b}{a\minus{}2b}\equal{}5$?

The given implies that $ 4a\plus{}3b\equal{}5a\minus{}10b\implies a\equal{}13b$, so substituting this into the desired expression gives $ \frac{a\plus{}11b}{a\minus{}b}\equal{}\frac{24b}{12b}\equal{}\boxed2$.


Suppose $a,b,c \in {\mathbb R }$ and denote  $\Delta=3\sum a^2 -(\sum a )^2 \; .$     [color=green][b] Prove that
\[ \begin{array}{|c|} \hline \displaystyle a+b+c + \sqrt{\displaystyle \frac{\Delta}{2}} \le 3 \cdot \max\{a,b,c\} \le \displaystyle a+b+c +\sqrt{2\Delta}\\ \hline \end{array} \; . \]
[/b][/color]
[size=75][b]Remark:[/b][color=blue][b] Assume   $S: =\sum a$ and $\Delta$  as known.  Then at the left the equality holds if and only if    the numbers  $\{a,b,c\}$are
\[ \frac{ S-\sqrt{2\Delta}}{3} ,\frac{ S-\sqrt{2\Delta}}{3} ,\frac{ S+\sqrt{\frac{\Delta}{2}}}{3} .  \]
At the right, the equality is valid only for the numbers  $\frac{S-\sqrt{\frac{\Delta}{2}}} {3} , \frac{ S-\sqrt{\frac{\Delta}{2}}}{3} , \frac{ S+\sqrt{2\Delta}}{3} \; .$

[/b][/color][/size]

Solution:
Wlog:$a \ge b \ge c$
$Left$
then the left of the inequality  $\displaystyle a+b+c + \sqrt{\displaystyle \frac{\Delta}{2}} \le 3 \cdot \max\left\{a,b,c\right\}$
$\iff \displaystyle a+b+c + \sqrt{\displaystyle \frac{\Delta}{2}} \le 3a$
$\iff \displaystyle b+c + \sqrt{\displaystyle \frac{\Delta}{2}} \le 2a$
$\iff \displaystyle \sqrt{\displaystyle \frac{\Delta}{2}} \le (a-b)+(a-c)$
$\iff \displaystyle \sqrt{\displaystyle \frac{\Delta}{2}} \le (a-b)^2+(a-c)^2+2(a-b)(a-c)$
$\iff (a-b)^2+(b-c)^2+(c-a)^2 \le 2(a-b)^2+2(a-c)^2+4(a-b)(a-c)$
$\iff (b-c)^2 \le (a-b)^2+(a-c)^2+4(a-b)(a-c)$
but we have:
$\iff (b-c)^2 \le (a-c)^2$
then the last inequality is obvious true
the equality is hold for $a=b$
which satisfy the remark you wrote.
$Right$
$3 \cdot \max\left\{a,b,c\right\} \le \displaystyle a+b+c +\sqrt{2\Delta}$
$\iff 3a \le \displaystyle a+b+c +\sqrt{2\Delta}$
$\iff 2a \le \displaystyle b+c +\sqrt{2\Delta}$
$\iff (a-b)+(a-c) \le \displaystyle \sqrt{2\Delta}$
$\iff (a-b)^2+(a-c)^2+2(a-b)(a-c) \le 2(a-b)^2+2(b-c)^2+2(c-a)^2$
$\iff 2(a-b)(a-c) \le (a-b)^2+2(b-c)^2+(c-a)^2$
but we have:$2(a-b)(a-c) \le (a-b)^2+(c-a)^2$
then the last inequality is obvious true
the equality is hold for $b=c$
which satisfy the remark you wrote.

The inequality is proved :)

[quote="zhaobin"]Solution:.....The inequality is proved.....[/quote] Nice solution !

$\begin{array}{|c|} \hline \displaystyle a+b+c - \sqrt{\displaystyle {2\Delta}} \le 3 \cdot \min\{a,b,c\} \le \displaystyle a+b+c -\sqrt{\displaystyle \frac{\Delta}{2}} \\ \hline \end{array} \; .$
is still true,can prove by same way ;)


conics ellipse geometry circumcircle ratio

Find the minimum possible area of an ellipse which encloses a 3,4,5
right triangle.
 :maybe:

This is POW, so I think that should be locked, but, I would like to give him alternative problem, so let's don't reply the answer to the problem.

Anyway the alternative problem:

[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=17673&search_id=190171756]2000 Tokyo University entrance exam/Science[/url]

:maybe:

kunny ( or anyone )

can you tell me if im on the right track?

Let us first find the answer for a regular triangle with area 1. Let the side of this triangle be a, and the radius of its circumcircle r. It is pretty clear that the circumcircle is the ellipse of minimum area for this triangle. So, let us find a first. The formula for the area of a regular triangle is:

S = a^2 * sqrt(3)/4
1 = a^2 * sqrt(3)/4.
a^2 = 4/sqrt(3)
a = 2/(3^(1/4))

Then, r = a/sqrt(3), so
r = 2/(3^(3/4))

Then the area of the circumcircle is
x = pi*r^2 = pi*4/(3^(3/2)) = pi*4/3/sqrt(3).

Now, back to your problem. Suppose that you've found such an ellipse. Then let us do an affine transform of the plane that will map your 3,4,5 right triangle into the regular triangle with area 1. Clearly, the ellipse of minimal area will map into the ellipse of a minimal area that encloses the new triangle, and we know that it is the circumcircle.

Now, when we do an affine transform, the ratios of all areas stay the same. In the "image" picture we have a regular triangle with area 1 and the minimal circle around it with area x = pi*4/3/sqrt(3). The ratio is
pi*4/3/sqrt(3). This means that on the original picture the ratio was the same. The area of the original triangle is 3*4/2 = 6, so the area of the ellipse we need is 6*pi*4/3/sqrt(3) = 8pi/sqrt(3). This is the answer.
  :|

mathzoid, what't the source of your problem? :mad:

This should be locked.

Japanese Communities Moderator

kunny

:maybe:

Its a problem from a friend who asked me about it.

is the solution correct?

The original problem is Here!

[url]http://www.math.purdue.edu/pow/fall2008/pdf/problem8.pdf[/url]

I think you would be able to understand why this problem should be locked.

oh ...

you kept saying "POW" ...that means prisoner of war!  :maybe:

I thought you where calling me names!!

now I get why u where upset since this is a problem of the week .

I will tell the person who emailed it to me that I wont help

thanks
 :ninja:

see my other post to your problem

you may wish to give me one like that

 :ninja:


FJI=108
so FJA=AJF=72.

180-72*2=36

answer : 36

who is posting all these probs


Do you know who won the science bowl in NC?

East Chapel Hill for the third year in a row

I hope we go against you. I haven't done well in math contests this year, but I will unleash my fury on [b]anybody[/b] that faces us.

yeah, east chapel hill won - but ENLOE had both teams in the top 4.

it could have been 2 teams in the top 3 - I think that our B team could have handled NCSSM A.

but with east having an IMOer, an IBOer, and an IPhOer (the last one is by hearsay), Enloe's teams stood no chance.

[quote="Teki-Teki"]

but with east having an IMOer, an IBOer, and an IPhOer (the last one is by hearsay), Enloe's teams stood no chance.[/quote]

damn

impressive, but still can't see due to my speed.

...?

that made very little sense, if any

[quote="mdk"]still can't see due to my speed.[/quote]

EH?

[quote="mdk"]Good Job East Chapell hill for winning third place. The game on stage was pretty good and you guys almost won.

MysticTerminator, I saw this guy throw your shot twice into bushes. What happened ? You looked intimitated. On the other hand the other guy on your team, Varun was pretty good.[/quote]

so east took third national?  nice job if so...

what is this 'game on stage'?

what do you mean by 'throw your show twice into bushes'?

dude I suspect he is just trying to belittle me

which is kinda stupid

since dude I am awesome

You see the top three teams get to play their games in front of everyone.

Also I meant his shot. Who was that guy Mystic Terminator?

Also, why would I try to belittle you? you don't have to be good at anything.

dude are you aware that you make [i]no sense[/i]

correction.

You see the top three teams get to play their games in front of everyone.

Also I meant his shot. Who was that guy Mystic Terminator?

Also, why would I try to belittle you? you don't have to be good at everything.

dude are you aware that you make [i]no sense at all[/i]

Um...all right then.

Did anyone notice, all the posts are kind of rated low...

Hmm, let us consider why  :roll:

Anyway, either the discussion needs to stay more on topic and less spam, or I will lock the topic.

[color=red]b-flat[/color]

Well, I did have one question related to the topic...

has anyone from NC (particularly East Chapel Hill, since they won it all with science bowl) met people from Parkview or Northview before? If so, could you say who? It'd be a bit interesting to know who you guys have gone against  :)

[quote="blueboat"]b-flat,

i don't think people will stop spamming unless you lock this topic, so i advise you to just lock it..[/quote]

Probably true; I'm leaving it open for a little bit to see if anyone answers Karth's question (I'm also deleting any spam that has happened since my last post).

b-flat, this thread is getting pretty silly and out of hand, and is an opening for DAVID BAY RUSH to spam in our forum besides. I advise you lock it.

we played parkview twice I think

I had met linda at MOP previously

which is why I followed them around yelling PARKVIEW

also hmm I don't recognize the other school name

Okay, I'm good, you can lock it now  :D Btw, I'm from Northview. We usually win at math tourneys, but Parkview usually wins at science bowl, I think.

[quote="Boy Soprano II"][quote="Karth"]We usually win at math tourneys[/quote]
[url=http://math.kennesaw.edu/~ckoppelm/Winners%202006-07]Whatever[/url] [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2007-12a/07-3-1012SchHR.shtml]you[/url] [url=http://www.unl.edu/amc/e-exams/e6-amc12/e6-1-12archive/2007-12a/07-4_03-1012SchMR.shtml]say[/url], Karthik.  Or did you just mean against Parkview?[/quote]

Dude I don't mean against your godly Lassiteriness. I mean just the Parkview people. Eh, but that was last year. But this year, we'll win it all  :wink:

EDIT: What the cheesecake? You guys had a combined 405 on AMCs this year?  :ninja: That's pretty monstrous. Nice job!

Thanks.  We got Mr. Slater an award for it. :D   You do have a real chance this year, though, unless somebody pops out of the woodwork on our behalf (not that I haven't been trying to make that happen).

Heh, we have a secret weapon that nobody can even dream of touching. He will be on JV next year, but he will be able to take on half the varsity in our school.  :evilgrin:

Karth, Sitan isn't a secret (at least to me). But why is this in the North Carolina forum? The North Carolina folks must be getting tired of it by not. (But at least it's not just Arnav bashing!)

It's not Sitan  :| Though that's a good guess... he'll be amazing.  :D

Eh, I don't know how this got into NC... it was mainly:

Science Bowl >> Math Tourneys >> Lassiter winning >> Northview winning ? >> because of secret weapon

There is no more on topic discussion, so this is now locked.


AMC
USA(J)MO
USAMO
combinatorics unsolved
combinatorics

[i]The $ n$-element subsets of a set of size $ n^2\plus{}n\minus{}1$ are distributed into two groups. Prove that one of those groups contains $ n$ pairwise disjoint sets. [/i]

Actually USAMO 2007: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=825499#p825499


quadratics
induction
number theory unsolved
number theory

Let $n$ be a positive integer.Prove that the following two statments are equivalent.

1.$n$ is not divisible by $4$.

2.There exist $a,b\in{Z}$ such that $a^2+b^2+1$ is divisible by $n$.

1) $n$ is not divisible by 4 means $n \not \equiv 0 \mod 4$

2) if $n$ would be divisible by 4, $n \equiv 0 \mod 4$, so $a^2+b^2+1$ should be $\equiv 0 \mod 4$, but this is impossible, since quadratic residual mod 4 are 0,1

Don't know if there is an elegant way for 2. $\implies$ 1. but:

Lemma 1: for $p$ a prime, we can find $x,y$ with $x^2+y^2+1 \equiv 0 \mod p$.
Since $p=2$ is stupid and annoying, lets assume $p$ odd.
Look at the sets $S : = {0^2 , 1^2 , 2^2 , 3^2 , ... , ( \frac{p-1}{2} )^2 }$ and $T: = {-1+0^2 , -1- 1^2 , -1-2^2 , -1-3^2 , ... , -1- ( \frac{p-1}{2} )^2 }$.
By a simple calculation we see that no element occurs twice in $S$ or $T$ when seen $\mod p$. But since both together have $p+1$ elements, there is $c \in S$ and $d \in T$ with $c \equiv d \mod p$, thus there are $x,y$ with $c=x^2 , d=-1-y^2$ and the result follows.

Lemma 1: for $p$ an odd prime and $s$ a positive integer, we can find $x,y$ with $x^2+y^2+1 \equiv 0 \mod p^s$ and $p \nmid x$.
Proof by induction on $s$:
Case $s=1$ is clear by Lemma 1 (when $p|x$ we change $x \leftrightarrow y$).
For the general one, take $x,y$ with $x^2+y^2+1 \equiv 0 \mod p^{s-1}$ and $p \nmid x$. Then we get that $(x+c p^{s-1})^2 \equiv x^2 + 2xc p^{s-1} \mod p^s$, giving when choosing $c$ such that $2xc p^{s-1} \equiv x^2+y^2+1 \mod p^s \iff 2xc \equiv - \frac{x^2+y^2+1}{p^{s-1}} \mod p$ the result (with $z=x+cp^{s-1},y$ the new pair). But such $c$ clearly exists since $p$ is odd and $z$ is not divisible by $p$ then.

Now by factoring $n$ and solving $x^2+y^2+1 \equiv 0 \mod p^s$ for every prime-power $p^s$ dividing $s$, we get by the Chinese Remainder Theorem $a,b$ with $n|a^2+b^2+1$.

Edit: see below for what was not correct.

[quote="ZetaX"]Then we get that $(x+c p^{s-1})^2 \equiv x^2 + 2c p^{s-1} \mod p^s$ [/quote]
Why?
I think it should be $(x+c p^{s-1})^2 \equiv x^2 + 2xc p^{s-1} \mod p^s$
and then it seams to me that your solution doesn't work.

Yes, the $x$ is missing, but it still works (it's clear that WLOG $p\nmid x$ when we add that to the induction).


geometry
3D geometry
prism
function
ratio
calculus

Me done the worst so wont say anything  :( .
Just want to say that maybe the 90 percentile cut off will be 130  :maybe:

[quote="pardesi"]not done well expecting around 210($ \pm 3000%
$) after my last part tests checking[/quote]

Dei i will murder u if u start behaving like Sachin  :mad: . Vetti scene potteinna aruvala edutthu seeviduvein sollitein  :P

sachin(film or cricketer  :P ) what scene...i know of   guy expecting 230...and many more whom i don't know would be expecting much more...
hope they do correct some of the answers ..

Yes their solns are much more pathetic than ever   :censored:  :mad: . And even if the topper gets 450 i dont care you are still putting vetti scene  :P

hey that will reduce...i saw now that my YDSE according to them is wrong...ofcours ethe fring width for red is twice as that of violet...what have they given...then the prism para 's no.1 question....
then the second match the following in maths when $ b>0$ then it can have i solution also.
taken$ x^{2}\plus{}2x\plus{}1$

expecting  abt 200. :P

.thre was a sum copied from math olympiad treasures (andreescu) :D . :moose:

Expecting 187 but their solutions are the most pathetic than ever :furious:  :bomb:  :censored: . They have blabbered in chemistry the most I would like to post a query to them saying them to change it. Anyway chapli is getting 227

f:[1,13}->R ,convex,.

integral(1,3)+integral(11,13)>=integral(5,9)

there on chapter on convex functions

[quote="pardesi"]hey that will reduce...i saw now that my YDSE according to them is wrong...ofcours ethe fring width for red is twice as that of violet...what have they given...then the prism para 's no.1 question....
then the second match the following in maths when $ b > 0$ then it can have i solution also.
taken$ x^{2} \plus{} 2x \plus{} 1$[/quote]

And 8th ques in Chemistry abt the eqvt weight of $ \ce{H3PO2}$ the ans shd be M/4 and not M/2  :mad:

apidiya i took that correct  :rotfl:
so -6 :rotfl:  :rotfl:  :rotfl:

[quote="shruthibaskaran"]And in one of the chem matches - graphite n diamond have heterogenous equilibrium?
i am really confused now..... the paper was quite vague actually....[/quote]

And so was for the one btwn rhombic and monoclinic sulphur  :mad:   :D . Btw how much are you expecting?  :maybe:

Hey folks,
Im gettin 229.. Know its kinda pathetic.. But jus dusn matter to me..

@Chapplikka: Pathetic da. :(

[quote="madness"]One question  :D . I just got FT1 and 2. Why the hell did he give us the answer sheets like that? Are we supposed to send it to him?  :P . Or are we supposed to install a software for evaluating that?  :rotfl:[/quote]

How dumb can you get?? You are supposed to self-correct it, but ofcourse if you want the stupid software used by FIITJEE, I can program it and send, wouldn't be too tough. :ROFL

Anyway post ur FT I scores if you have written it.

From what I know, Chapplikka has got 240+- in FTI and Physici Olympiaci has got 134 in Paper I.

[quote="Cyber-Shreyas"]@Chapplikka: Pathetic da. :(

[quote="madness"]One question  :D . I just got FT1 and 2. Why the hell did he give us the answer sheets like that? Are we supposed to send it to him?  :P . Or are we supposed to install a software for evaluating that?  :rotfl:[/quote]

How dumb can you get?? You are supposed to self-correct it, but ofcourse if you want the stupid software used by FIITJEE, I can program it and send, wouldn't be too tough. :ROFL

Anyway post ur FT I scores if you have written it.

From what I know, Chapplikka has got 240+- and Physici Olympiaci has got 134 in Paper I.[/quote]

Noo i was wondering why he had given that kind of sheets for us to write in??? Maybe for practise though bcos we might as well write in plain sheets and correct for ourselves which would be easier  :D  :D

It will surely be easier to correct if we write in our style but its good practise for JEE the other way. He has a beautiful point in his letter. Write the test only during JEE timings and not night to tune your body clock. Night Birds like me need to change. :(

Pardesi, I recieved almost a week ago.

[quote="pardesi"]yes thats a nice point shreyas i read it a long time ago[/quote]

I find it much more convenient to work in the night time

1) No disturbance from Other Family Members
2) No Phone Calls (except random messages from chapplikka :) )
3) Peace
4) Can take breaks for 5-10 mins randomly, watch TV (usually TV isnt' free before 10 in my house) and sneak into the kitchen and eat biscuits

And there is something special about the ambience in the night that encourages me to study in the night.

Disadvantages

1) Mosquito :D

yes add one more thing to advantages ....music.... :D
and for me subtract disadvantages here room temp.at night is around 18 degree celcius mosquitoes will die :D

[quote="Cyber-Shreyas"][quote="madness"]my advantages: Same as Shreyas and add to it access to the comp and aops  :D
disadvantage: If i lie down and read, I sleep but then i wanna lie down and read  :P[/quote]

Dunno if this is ad or disad.. but nevertheless I have no access to computer at night. :( unless I have some very important job like C++ Project. :)

And Pardesi, is it that cold in Bhubaneshwar??[/quote]

Thats an adv to me bcos if i am on the comp i wont sleep easily otw i might study in the middle of naps  :D

in FT1 ,did paper1 well
but flopped very badly in paper2
i am getting only 192..................
paper1-128
paper2-64

[quote="shruthibaskaran"]Rhomic and monoclinic sulphur will have heterogenous eq as one of them is in liquid form, not sure about which one.. But diamond and graphite? God this sucks...

Havent totalled it yet... It crossed 100 as far as I know, will total and post here by tomorrow probably.... What bt ya Rohit?

And yea, the fringewidth question is also wrong..... :( This is pathetic man....[/quote]

rhombic and monoclinic are Bravais lattices so they are necessarily solids...please explain how is it possible to have either of rhombic and monoclinic sulphur in liquid form??????????

If sec a + cosec a = p has four solns in (0,2pi) then PT. p^2>8.  ......from fiitjee supple pack.......

next ques is same for only 2 soln in the given ques....but PT. p^2<8 :?:

and Here room temp at 3.00 am is about 2.6 degrees.......leave alone the mosquitoes........I myself am dying :wink:


induction
limit
function
real analysis
real analysis unsolved

let  $ x_0 \in (0,1]$be real number  and $ u_n,x_n$ are two sequences such that ;
$ u_n\equal{}[\frac{1}{x_n}]\plus{}1$,and $ x_{n\plus{}1}\equal{}u_nx_n\minus{}1.$
prove$ \frac{1}{u_0}\plus{}\frac{1}{u_0u_1}\plus{}....\frac{1}{u_0u_1u_2...u_n}$ converge to $ x_0$

Well, i could prove this for rational numbers: it is easier because the sequence stabilizes and this allows to use induction. Here it goes:
1) Suppose $ x_0 \equal{} \frac1q$. Then $ u_0 \equal{} q \plus{} 1, \ x_1 \equal{} \frac1q, \ldots$, $ u_k \equal{} q \plus{} 1 \ \forall k$. Then our sum is exactly
$ \sum\frac {1}{(1 \plus{} q)^k} \equal{} \frac1q$.
2) Let $ x_0 \equal{} \frac p q$. If $ q \equal{} q_1p \plus{} r_1, \ r_1 < p$, then $ u_0 \equal{} q_1 \plus{} 1$, $ x_1 \equal{} \frac {p \minus{} r_1}{q}$ ; $ q \equal{} q_2(p \minus{} r_1) \plus{} r_2$, then $ u_1 \equal{} q_2 \plus{} 1, \ x_2 \equal{} \frac {p \minus{} r_1 \minus{} r_2}{q}$. So, as far as $ x_n > 0$, our sequence must stabilize at most in $ (p \minus{} 1)$ steps on the value $ \frac1q$. We shall use induction on the number $ N$ of steps to prove the statement.
Case 1) shows that $ N \equal{} 1$ holds. Suppose that statement is true for $ N < k$, and let's investigate $ N \equal{} k$.
Taking $ x_1$ as a starting point, we see that sequence will stabilize in $ N \minus{} 1$ steps, so the sum
$ \frac {1}{u_1} \plus{} \frac {1}{u_1u_2} \plus{} \dots \plus{} \frac {1}{u_1\dots u_m} \plus{} \dots \equal{} x_1 \equal{} \frac {p \minus{} r_1}{q}$.
Then

$ \frac {1}{u_0} \plus{} \frac {1}{u_0u_1} \plus{} \dots \plus{} \frac {1}{u_0\dots u_m} \plus{} \dots \equal{} \frac {1}{u_0}(1 \plus{} x_1) \equal{} \left(\frac {p \minus{} r_1}{q} \plus{} 1\right)\frac {p}{q \plus{} p \minus{} r_1} \equal{} \frac pq$

For now, i dont see the trick which allows to generalize this for irrational $ x_0$.

I may be mistaken but, if:

(1)
$ {u[n]\equal{}\frac{1}{x[n]}\plus{}1}$

and

(2)
$ {x[n\plus{}1]\equal{}u[n]x[n]\minus{}1}$

then plugging (1) into (2) we get:

$ {x[n\plus{}1]\equal{}(\frac{1}{x[n]}\plus{}1)x[n]\minus{}1}$
$ {x[n\plus{}1]\equal{}x[n]\plus{}1\minus{}1}$
$ {x[n\plus{}1]\equal{}x[n]\equal{}C}$

Now:

$ {u[n]\equal{}\frac{1}{C}\plus{}1\equal{}\frac{C\plus{}1}{C}}$

The sum we have at the end can then be represented as:

$ {S\equal{}\frac{C}{C\plus{}1}\plus{}(\frac{C}{C\plus{}1})^2\plus{}(\frac{C}{C\plus{}1})^3...\plus{}(\frac{C}{C\plus{}1})^n\equal{}\frac{\frac{C}{C\plus{}1}^{n\plus{}1}\minus{}1}{\frac{C}{C\plus{}1}\minus{}1}\minus{}1}$

and since $ {C\equal{}x_0\in(0,1]}$

$ {\lim_{n\rightarrow\infty}\frac{\frac{C}{C\plus{}1}^{n\plus{}1}\minus{}1}{\frac{C}{C\plus{}1}\minus{}1}\minus{}1\equal{}\minus{}\frac{1}{\frac{C}{C\plus{}1}\minus{}1}\minus{}1\equal{}C\equal{}x_0}$

Unless, $ {[x]}$ denotes the fraction part in witch case I'm way off!

[quote="milin"]
Unless, $ {[x]}$ denotes the fraction part in witch case I'm way off![/quote]
$ [t]$ denotes the biggest integer not bigger than $ t$. :)

[quote="milin"]Ah! I'm way of in that case as well! :)[/quote]
Try to read previous messages before posting yours...

And to complete the proof for irrational numbers, one can use the fact that for any $ x$ we can choose $ p,q: \ \mid x \minus{} \frac pq\mid < \frac {1}{q^2}$, and $ p,q$ can be made arbitrary large. Then, we can prove(for example, by induction), that for any $ n$ we can choose $ p_n, q_n$ such that $ \{u_0(x),\dots, u_n(x)\} \equal{} \{u_0(\frac {p_n}{q_n}),\dots,u_n(\frac {p_n}{q_n})\}$, from which the conclusion follows immediately.

An other proof ... :D
We denote the functions $ X_n: ]0,1]\rightarrow R ,\  U_n: ]0,1]\rightarrow R$ such that $ X_0(t)\equal{}t ,\  U_n(t)\equal{}[\frac 1{X_n(t)}]\plus{}1 , \ X_{n\plus{}1}(t)\equal{} X_n(t)U_n(t)\minus{}1$.
It's easy to see that $ 0 < X_{n\plus{}1}(t)\leq X_n(t)\leq t$ because $ \frac 1{X_n(t)} < U_n(t) \leq \frac 1{X_n(t)}\plus{}1$ ,  so $ U_{n\plus{}1}(t)\geq U_n(t)\geq 2$
We  denote $ s: ]0,1]\rightarrow R$ such that $ s(t)\equal{}\sum_{m\equal{}0}^{\infty}1/(\prod _{k\equal{}0}^m U_k(t))$. We have $ \frac 1{U_0(t)} < s(t) \leq \sum_{m\equal{}0}^{\infty}\frac 1{U_k^m(t)}\equal{}\frac 1{U_0(t)\minus{}1}$, so $ U_0(s(t))\equal{}U_0(t)$.
We  prove by induction that $ X_n(s(t))\equal{}\sum_{m\equal{}n}^{\infty}1/(\prod _{k\equal{}n}^m U_k(t)) , \  U_n(s(t))\equal{}U_n(t)$, so $ s(s(t))\equal{}s(t)$.  Now we will prove that the function $ s$ is injective
If  $ U_n(a)$ and $ U_n(b)$ are not identical then  there is $ m$ such that $ \forall _{0\leq k<m}U_k(a)\equal{}U_k(b),\ U_m(a)\neq U_m(b)$, for ex. $ U_m(a) < U_m(b)$ .
If $ U_m(a) < U_m(b)$ then $ U_m(a) \leq  U_m(b)\minus{}1$, so  $ s(b)< s(a)$ because
$ s(b)\minus{}s(a) < \frac 1{U_0U_1...U_{m\minus{}1}}((\sum _{i\equal{}1}^{\infty}\frac 1{U_m^i(b)})\minus{}\frac 1 {U_m(a)})\equal{}$ $ \frac 1{U_0U_1...U_{m\minus{}1}}(\frac 1{U_m(b)\minus{}1}\minus{}\frac 1 {U_m(a)})\leq 0$
Let $ a,b$ be such that $ s(a)\equal{}s(b)$ , so $ U_n(a)$ and $ U_n(b)$ must be identical , and we can write :
$ |X_{n\plus{}1}(a)\minus{}X_{n\plus{}1}(b)| \equal{} |X_n(a)\minus{}X_n(b)|.U_n \equal{}$$ ... \equal{} |(X_0(a)\minus{}X_0(b)|.U_n...U_0 \geq |a\minus{}b|.2^{n\plus{}1}$
But $ X_{n\plus{}1}$ is bounded , so it's necessary to have $ |a\minus{}b|\equal{}0$, so $ s$ is injective.
Finally from $ s(s(t))\equal{}s(t)$, $ s$ injective,   we obtain $ s(t)\equal{}t$ (q.e.d.)
 :cool:


I'm not sure if this topic relates to this forum, but since it's very quick, can someone answer it? (it has a little to do with math, like modulus...)

This partial program is part of a bigger program, but essentially I want to take a modulus and a power (so I want to check all numbers 0, 1, 2, 3, ..., m-1 taken to a power p and then take it modulus m - BTW is there a shortcut key on TI 84 Silver Edition that lets you check numbers 0, 1, 2, 3, ... m-1 for a specific command or do you have to use repeat? And you use a repeat ... repeat... end... end (so a repeat within a repeat? How would you do that?)

Here is my partial program:

Prompt M,P
X=1
Y=0
Repeat 0 < (or equal sign) B (less than or equal sign) M-1
X^P store A
(A-(Y*M))store B
Y+1 store Y
End
Disp B

Yet when I run it, I get (for M=2 and P=3, 68261289.04, which clearly isn't right (in a modulus, you're supposed to get an integer less than M)).

Can anyone tell me how to fix my program??????

anyone?

or can someone recommend me a forum to place this in?

hmm don't know the right forum, but you can also use the For function.
If you use
For([i]var, A, B, [Increment][/i]) notation from TI CtlgHelp
[i]commands[/i]
End
The program takes the variable, first executes [i]commands[/i] with the variable equalling A, adds to the variable the increment (1 by default), and repeats until the variable is greater than B.
There are tricks you can use with For(, but that's the basics
example:
For(A,1,5)
Disp A
End

Result:
1
2
3
4
5

as for your remainder problem, i recommend , for finding the remainder of N divided by M,
MfPart(N/M)

sorry, but where do you find the command MfPart(N/M)

(I could not see it in the catalog :( )

I think that's M * fPart( N / M ). fPart (or something similar) gives the non-integer part of the quotient.


Quantum of Light and Life.

Question.
What inhale Life in the formulas and in the equations ?
What must be present in a body to make it alive ?
The answer.
Soul. Quantum of Light.
*         *         *
According to Pauli Principle only one single electron can be in the atom.
This electron recovers the atom.
If the atom containing more than one electron (for example – two),
this atom represents “ Siam twin”.
Rescue us the God to have such children.
Each of us has in himself quantum of light – electron.
But we do not know of this.
Whether and it is possible to find out it?
Yes. It is possible.
The theory of knowledge gives the positive answer.
*        *       *
The theory of knowledge.
The man acts: 1) at usual under logic program,
                       2) sometimes on intuition (unconsciousness).
*             *            *
If you have time and desire, I ask you to visit my site
http://www.socratus.com
  Best wishes.
Socratus.

If Valentin or someone would like to start a philosophy forum, this would belong [i]there[/i]. But in this forum, please try to ask questions or make comments that pertain to the biological sciences.


In a school of 250 students, everyone takes one English class and one history class
each year. Today, 15 total students were absent from their English class and ten total
students were absent from their history class. Five of the students were absent from
both classes. If a student is chosen at random from this school, what is the probability
that s/he was not absent from either class? Express your answer as a percent.

[hide="Solution"]
I don't know how to draw Venn Diagrams in Latex / Asymptote, but, have one circle represent students absent from history, one circle represent students absent from english, and their intersection represent students absent from both. It is given that 5 students were absent from both, so label the intersection 5. We know the total students absent from english was 15, so $ 15 \minus{} 5 \equal{} 10$ students were only absent from english. Also, 10 students were absent from history, so $ 10 \minus{} 5 \equal{} 5$ students were only absent from history. Therefore, exactly $ 10 \plus{} 5 \plus{} 5 \equal{} 20$ students were absent from at least one class. So there are 230 students out of 250 who were not absent from either class, so the probability of picking one of them is $ \frac{230}{250} \Rightarrow 92$ percent [/hide]

15-5=10 people were missing from English class only today, and 10-5=5 people were missing from history class only, so a total of 10+5+5=20 people were missing from school today. The chance of calling someone who was in class today is 230/250=92%. Of course, the chance of calling somebody in class today would be 100% or else the teacher wouldn't call on them.


A game start with a pile of 25 markers. Two players alternatively remove 1,2 or 3 of them. When all the markers are taken, the winner is the player who has an even number of markers. Which player has a winning strategy?

Pierre.

Consider the game $G_i$ with $i$ markers: If first has even number of sticks he wins. Also consider the reversed $R_i$ game where if first has odd sticks he wins :
Denote $a_i=0$ if one wins in game $G_i$ and $a_i=1$ else. Analogously define $b_i$ for $R_i$.
One computes $a_1,a_2,a_2,b_1,b_2,b_3$ and now he can easily express reccurently:$a_k,b_k$ through the previous terms:if first takes one,two,or three sticks we get three new games and first wins iff at least one of the is losing.
Now one shall just compute $a_{25}$

[quote="pbornsztein"]Hmm... I was hoping for something else... :(

Pierre.[/quote]
I too.

I don't understand something, because as Iura says, $G_2$ and $G_4$ and $R_4$ and $R_2$ are draws!!!!!!! I think some extensions are necessary...

I think that we must also define a type of draw, that is, does our draw game end with both players having an odd or even number of stones....
But also, 4 will give us an odd draw for sure, but also a possibility to have an even draw....Although the 1st player can ensure an odd draw....
In this case, we can say that 1 is defintely loosing for the 1st player,
2 is a possibility for the 1st player to choose either an even or odd draw.
3 is winning for the 1st one, loosing for the second one.
4 is a very big problem, because the first player can only ensure an odd draw, by taking 3. But he cannot ensure an even draw......
Also, the second player can ensure an odd draw...
For 5, it is winning for the 1st one, as he can take one stone, and because he will be the second player in the game with 4 stones, he can ensure an odd draw in it!!!
For 6, thing are becoming more difficult...I think I know the possibilities, but they are hard to explain.
Plus, I now realize that we must also keep for every position if the first or second player can get to (odd,even) or (even,odd).
I don't think now that this is the approach. Sorry for the big message, at the beggining I thought something is possible, but now I realize it is not so!! Sorry :blush:

Ok, to say some of my thoughts about this problem:
First of all, let's see that the player that will move in the position with 2 or 3 stones will definetly win!!!! In the position with 4 stones, he will win only if he has an odd number of stones. For 5 stones, he will win only if the second player has an even number of stones....
All this comes from the fact that 2 and 3 are winning possitions for the player at move.....


A sports stadium is circular with radius 60m, surrounded on all sides by buildings height 10m, what is minimum speed with which a ball must be kicked from the centre to completely clear the stadium.

Any suggestions?

[hide]
do it using position vectors
$ \tilde{a} = - 9.8j$
$ \tilde{v(t)} = ai + (b - 9.8t)j$
$ \tilde{r(t)} = ati + (bt - 9.8t^2)j$

We try to minimise $ |v(0)| = \sqrt {a^2 + b^2}$ given that it passes through $ (60,10)$.

so $ 60 = at$ and $ 10 = bt - 9.8t^2$

So $ |v(0)| = \sqrt {\left(\dfrac{60}{t}\right)^2 + \left( \dfrac{10 + 9.8t^2}{t} \right)^2}$

We can find the min without the squareroot to make it easier
After some calculus
$ t = 2.49137...$

$ v_{min} = 14\sqrt {\sqrt {37} + 1}$

[/hide]

EDIT: Yes, it should be 4.9 instead of 9.8  :oops: . It's extremely exacerbating when you make a mistake like that at the very begining of the question. But, as Math154 demonstated this is a pretty messy solution.

[quote="ocha"]

 $ 10 \equal{} bt \minus{} 9.8t^2$

[/quote]

would this not be

 $ 10 \equal{} bt \minus{} 4.9t^2$?

Also how did you get $ t$?

@JasonSullivan: Yes, I'm pretty sure it's $ 10 \equal{} bt \minus{} 4.9t^2$.

Making minor adjustments,

$ |v(0)| \equal{} \sqrt {\left(\frac {60}t\right)^2 \plus{} \left(\frac {10 \plus{} 4.9t^2}t\right)^2} \equal{} \sqrt {\frac {3600}{t^2} \plus{} \frac {100 \plus{} 98t^2 \plus{} 24.01t^4}{t^2}}$

$ \equal{} \sqrt {\frac {3700 \plus{} 98t^2 \plus{} 24.01t^4}{t^2}} \equal{} \sqrt {\frac {3700}{t^2} \plus{} 24.01t^2 \plus{} 98}$.

[hide="Non-Calculus Solution"]
By AM-GM, $ \frac {3700}{t^2} \plus{} 24.01t^2\geq2\sqrt {\frac {3700}{t^2}\cdot24.01t^2} \equal{} 2\sqrt {88837} \equal{} 98\sqrt {37}$,

which is achieved when $ \frac {3700}{t^2} \equal{} 24.01t^2\implies t^4\equal{}\frac{370000}{2401}\implies t \equal{} \frac {10\sqrt [4]{37}}7$.

Therefore, $ \sqrt {\frac {3700}{t^2} \plus{} 24.01t^2 \plus{} 98}\geq\sqrt {98 \plus{} 98\sqrt {37}} \equal{} \sqrt {49}\cdot\sqrt {2 \plus{} 2\sqrt {37}} \equal{} 7\sqrt {2 \plus{} 2\sqrt {37}}$.

We can let $ \sqrt a \plus{} \sqrt b \equal{} \sqrt {2 \plus{} 2\sqrt {37}}\implies a \plus{} b \plus{} 2\sqrt {ab} \equal{} 2 \plus{} 2\sqrt {37}$

$ \implies a \plus{} b \equal{} 2,ab \equal{} 37$, which has no solutions for $ a,b\in\mathbb R$.

Thus, our final answer is $ \boxed{7\sqrt {2 \plus{} 2\sqrt {37}}}$, which occurs when $ t \equal{} \frac {10\sqrt [4]{37}}7$.
[/hide]

[hide="Calculus Solution"]
The minimum of $ \frac{3700}{t^2}\plus{}24.01t^2$ occurs when the derivative of this is equal to zero, so

$ 48.02t\minus{}\frac{7400}{t^3}\equal{}0\implies 48.02t^4\minus{}7400\equal{}0\implies t^4\equal{}\frac{370000}{2401}\implies t \equal{} \frac {10\sqrt [4]{37}}7$, and the rest of the solution is similar to above.
[/hide]


ratio
geometry
geometric transformation
reflection
circumcircle
perpendicular bisector
cyclic quadrilateral

[quote="yetti"]$ G$ is centroid of a $ \triangle ABC,$ $ G_a$ its reflection in the perpendicular bisector of $ BC.$ Tangent of the triangle circumcircle $ (O)$ at $ A$ cuts $ BC$ at $ P$ and $ Q$ is midpoint of $ AP.$ Circles $ (P), (Q)$ with centers $ P, Q$ and radii $ PA, QA$ cut $ AG_a$ at $ U, V,$ respectively. Show that $ G_aU \equal{} 4 \cdot G_aV.$[/quote]
Maybe  I am silly, but just to be sure, can you check the problem again? I think you make a typo with your final statement

Yup, there is a typo. $ Q$ is not midpoint of $ AP,$ but midpoint of $ OP.$ Correct version:

$ G$ centroid of a $ \triangle ABC,$ $ G_a$ its reflection in the perpendicular bisector of $ BC.$ Tangent of the triangle circumcircle $ (O)$ at $ A$ cuts $ BC$ at $ P$ and $ Q$ is midpoint of $ \color{red}OP.$ Circles $ (P), (Q)$ with centers $ P, Q$ and radii $ PA, QA$ cut $ AG_a$ again at $ U, V,$ respectively. Show that $ G_aU \equal{} 4 \cdot G_aV.$

Let $ X$ be, the point of intersection, the other than $ A,$ of the circumcircle $ (O)$ of the given triangle $ \bigtriangleup ABC,$ from the circle $ (P),$ centered at $ P$ with radius $ PA.$

Because of $ PA\perp OA$ and $ PX\perp OX,$ we conclude that the circle $ (Q)$ with diameter $ OP,$ passes through the points $ A,\ X.$

We denote the points $ D,\ N,\ N',$ as the orthogonal projections of the points $ A,\ G,\ G'$ respectively, where $ G$ is the centroid of $ \bigtriangleup ABC$ and $ G'$ ( instead of $ G_{a}$ ) is the symmetric point of $ G,$ with respect to the line segment $ OM,$ where $ M$ is the midpoint of the side segment $ BC.$

$ \bullet$ From the cyclic quadrilateral $ AOMX$ we have that $ \angle OAX \equal{} \angle OMA$ $ ,(1)$ because of $ OA \equal{} OX$.

But, $ \angle OMA \equal{} \angle OMG'$ $ ,(2)$ because of the symmetry.

So, from $ (1),\ (2)$ $ \Longrightarrow$ $ \angle OAX \equal{} \angle OMG'$ $ ,(3)$

From $ (3)$ we conclude that the points $ G',\ M,\ X,$ are collinear.

$ \bullet$ Also, because of $ (3)$ and $ OM\perp MP,$ we conclude that the line segment $ MP,$ bisects the angle $ \angle AMX$ and then, we have $ \angle AMP \equal{} \angle PMX$ $ ,(4)$

From $ (4)$ we conclude that the point $ A'\equiv (P)\cap G'X,$ is the symmetric point of $ A,$ with respect to the sideline $ BC$ of $ \bigtriangleup ABC.$

$ \bullet$ From the power now, of the point $ G',$ with respect to the circles $ (Q),\ (P),$ we have  $ (G'V)\cdot (G'A) \equal{} (G'M)\cdot (G'X)$ $ ,(5)$ and  $ (G'A)\cdot (G'U) \equal{} (G'X)\cdot (G'A')$ $ ,(6)$ respectively, where $ V\equiv (Q)\cap G'A$ and $ U\equiv (P)\cap G'A$.

From $ (5),\ (6)$ $ \Longrightarrow$ $ \frac {G'V}{G'U} \equal{} \frac {G'M}{G'A'}$ $ ,(7)$

From $ G'N'\parallel AA'$, based on the [b][size=100]Thales theorem[/size][/b], we have that $ \frac {G'M}{MA'} \equal{} \frac {G'N'}{A'D} \equal{} \frac {GN}{AD} \equal{} \frac {1}{3}$ $ ,(8)$

From $ (8)$ $ \Longrightarrow$ $ \frac {G'M}{G'M \plus{} MA'} \equal{} \frac {1}{4}$ $ \Longrightarrow$ $ \frac {G'M}{G'A'} \equal{} \frac {1}{4}$ $ ,(9)$

From $ (7),\ (9)$ $ \Longrightarrow$ $ \frac {G'V}{G'U} \equal{} \frac {1}{4}$ $ ,(10)$ and the proof is completed.

$ \bullet$ This proof is dedicated to [b][size=100]Luis Gonzalez[/size][/b].

Kostas Vittas.

Thank you for the solution. This implies that power of the centroid $ G'$ to $ (P)$ is $ G'X \cdot G'A' \equal{} \frac {R^2 \minus{} OG^2}{AG} \cdot (AM \plus{} GM) \equal{} 2(R^2 \minus{} OG^2),$ where $ R$ is the circumradius. Power of $ O$ to $ (P)$ is equal to $ OA^2 \equal{} R^2.$ It follows that powers of $ O, G$ to the reflections of A-, B-, C-Apollonius circles in the perpendicular bisectors of $ BC, CA, AB$ are equal to $ R^2, 2(R^2 \minus{} OG^2),$ respectively, and they are coaxal with radical axis $ OG.$ (This was my 1st proof, before noticing reflections of the Apollonius circles of the medial triangle in its perpendicular bisectors.)


limit
logarithms
real analysis
real analysis unsolved

show the following:  :D

(you [u]cannot[/u] use complex variables, fourier series, and anything that is likely to be taught at the college / university level)

[ NOTE: $ p_k$ is the ${ k^{th}}$ prime number ... ]

for $ n\in\mathbb Z,\;\;n\ge 2$,
$ \text{(i)}\qquad \prod_{m = 1}^\infty \left(1 + \dfrac{1}{p_m^n}\right)\; = \;\boxed{\dfrac{\zeta(n)}{\zeta(2n)}}$

$ \text{(ii)}\qquad \lim_{a\to\infty}\;\left\{ \dfrac{\prod_{m = 1}^a \left(1 + \dfrac{1}{p_m}\right)} {\ln\,p_a}\;\right\} = \;\boxed{\dfrac{e^\gamma}{\zeta(2)}}$

( where $ \gamma$ is Euler's constant )

[hide="(i)"]
$ \prod_{m \equal{} 1}^\infty \left(1 \plus{} \dfrac{1}{p_m^n}\right) \equal{} \prod_{m \equal{} 1}^\infty \left( 1 \plus{} \frac {p_m^n \minus{} 1}{p_m^n (p_m^n \minus{} 1)} \right) \equal{} \prod_{m \equal{} 1}^\infty \left( \frac {p_m^{2n} \minus{} 1}{p_m^{2n} \minus{} p_m^n} \right) \equal{}$
$ \equal{} \prod_{m \equal{} 1}^\infty \left( \frac {p_m^{2n} \minus{} 1}{p_m^{2n}} \cdot \frac {p_m^n}{p_m^n \minus{} 1} \right) \equal{} \prod_{m \equal{} 1}^\infty \left( \frac {1}{1 \minus{} p_m^{ \minus{} n}} \cdot \frac { 1 \minus{} p_m^{ \minus{} 2n} }{1} \right) \equal{} \boxed{\dfrac{\zeta(n)}{\zeta(2n)}}$[/hide]

[hide="(ii)"]
use Mertens theorem
i.e. $ \lim_{n \to \infty} \frac {\prod_{k \equal{} 1}^n \left\{ 1 \minus{} p_k^{ \minus{} 1} \right\}^{\minus{}1}}{\ln p_n} \equal{} e^{\gamma}$
[/hide]

cool... i use the exact same method to show (i).

[quote="@petko"]
for (ii)...

use Mertens theorem
i.e. $ \lim_{n \to \infty} \frac {\prod_{k \equal{} 1}^n \left\{ 1 \minus{} p_k^{ \minus{} 1} \right\}^{ \minus{} 1}}{\ln p_n} \equal{} e^{\gamma}$
 [/quote]

sorry, ... no can do, merten's theorem is not known to 7th or 8th graders here in india.  :|

i think that pupils of the 7-th grade even don't know what is $ \ln x$, i'm not saying about $ \gamma$ and $ \zeta (x)$.  :D our pupils start studiing equations of the type $ x^2 \plus{} ax \plus{} b \equal{} 0$ in the 8-th grade when they are 14 years old.

[b]misan[/b] and what about sth like this:
$ \lim_{a\to\infty}\;\left\{ \dfrac{\prod_{m \equal{} 1}^a \left(1 \plus{} \dfrac{1}{p_m}\right)} {\ln\,p_a}\;\right\} \equal{} \frac{1}{\zeta (2)} \lim_{a \to \infty} \frac{\prod_m \left\{1\minus{} p_m^{\minus{}1}  \right\}^{\minus{}1}}{\ln p_a}$
And now use
$ \gamma \equal{} \lim_{x \to \infty} \left\{ \sum_{m \equal{} 1}^x \ln \frac{1}{1 \minus{} p_m^{\minus{}1}} \minus{} \underbrace{\ln \ln x}_{\equal{} \ln \ln p_x} \right\}$
and we're done.


combinatorics
algebra
Sequence
invariant
IMO Shortlist

Let $ R_1,R_2, \ldots$ be the family of finite sequences of positive integers defined by the following rules:  $ R_1 \equal{} (1),$ and if $ R_{n - 1} \equal{} (x_1, \ldots, x_s),$ then

\[ R_n \equal{} (1, 2, \ldots, x_1, 1, 2, \ldots, x_2, \ldots, 1, 2, \ldots, x_s, n).\]

For example, $ R_2 \equal{} (1, 2),$ $ R_3 \equal{} (1, 1, 2, 3),$ $ R_4 \equal{} (1, 1, 1, 2, 1, 2, 3, 4).$ Prove that if $ n > 1,$ then the $ k$th term from the left in $ R_n$ is equal to 1 if and only if the $ k$th term from the right in $ R_n$ is different from 1.

Some links, for convenience.

Thread with discussion: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=108138
OEIS: https://oeis.org/A004073

I think this is similar in flavor to the [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?f=41&t=47262]Russia 2003 Ana/Bora problem[/url], although that one might be a little trickier?

[hide="Solution"]Define the companion family of finite sequences $\{Q_{ni}\}$ such that $Q_{n1}=(1)$, $Q_{nn}=(n)$ and $Q_{ni}=Q_{(n-1)(i-1)}Q_{(n-1)i}$ whenever $1<i<n$.

[b]Claim 1.[/b] $Q_{ni}$ has length $\binom {n-1}{i-1}$.
[i]Proof.[/i] This is clear for $i=1$ and $i=n$. Using induction, we find that $Q_{(n+1)(i+1)}=Q_{ni}Q_{n(i+1)}$ has length
\[\binom{n-1}{i-1}+\binom{n-1}i=\binom ni,\]
as needed. $\blacksquare$

Now define for any finite sequence $A=(x_1,x_2,\dots, x_n)$ (where all the $x_i$ are positive integers) the following operation:
\[\hat{A}=(1,2,\dots,x_1,1,2,\dots,x_2,\dots,1,2,\dots,x_n).\]
Then of course we have $\hat{A}\hat{B}=\hat{AB}$.

[b]Claim 2.[/b] $Q_{ni}=\hat{Q_{(n-1)i}}$ for $n\ge 2$ and $1\le i\le n-1$.
[i]Proof.[/i] We use induction on $n$. This is trivial for $i=1$ because $Q_{n1}=Q_{(n-1)i}=(1)$. This also implies the truth of the claim for $n=2$. For $n>2$, $1<i<n$, we find by the induction hypothesis
\[Q_{ni}=Q_{(n-1)(i-1)}Q_{(n-1)i}=\hat{Q_{(n-2)(i-1)}}\hat{Q_{(n-2)i}}=\hat{Q_{(n-1)i}}.\blacksquare\]

[b]Claim 3.[/b] The $j$-th term from the left in $Q_{ni}$ is a $1$ if and only if the $j$-th term from the right in $Q_{n(n+1-i)}$ is not a $1$ ($n\ge 2$).
[i]Proof.[/i] Inducting on $n$, starting from the trivial $n=2$, we note that either $i=1$ or $i=n$, in which case we are done based on the definition, or else we have $Q_{ni}=Q_{(n-1)(i-1)}Q_{(n-1)i}$ and $Q_{n(n+1-i)}=Q_{(n-1)(n-i)}Q_{(n-1)(n+1-i)}$. In the latter case, the $j$-th term from the left in $Q_{ni}$ is either in the part $Q_{(n-1)(i-1)}$, or it is in $Q_{(n-1)i}$. Because of Claim 1, we notice that $Q_{(n-1)(n+1-i)}$ is just as long as $Q_{(n-1)(i-1)}$, so the $j$-th term from the left in $Q_{ni}$ and the $j$-th term from the right in $Q_{n(n+1-i)}$ coincide with either the $j$-th term from the left in $Q_{(n-1)(i-1)}$ and from the right in $Q_{(n-1)(n+i-1)}$, or the $j^*$-th term (for some $j^*$) from the left in $Q_{(n-1)i}$ and from the right in $Q_{(n-1)(n-i)}$. In both cases, we are done because of the induction hypothesis. $\blacksquare$

[b]Claim 4.[/b] $R_n=Q_{n1}Q_{n2}\dots Q_{nn}$.
[i]Proof.[/i] This is trivial for $n=1$. Proceeding by induction and using the definition of $R_n$ and also Claim 2, we see that
\[R_n=\hat{R_{n-1}}Q_{nn}=\hat{Q_{(n-1)1}}\hat{Q_{(n-1)2}}\dots\hat{Q_{(n-1)(n-1)}}Q_{nn}=\]
\[=Q_{n1}Q_{n2}\dots Q_{n(n-1)}Q_{nn}.\blacksquare\]

Now take the $k$-th term from the right in $R_n$. Using the identity of Claim 4, we see that this is inside some sequence $Q_{ni}$, at place number $j$ (from the left). Because of Claim 1, this implies
\[k=\binom {n-1}0+\binom{n-1}1+\dots+\binom{n-1}{i-2}+j=\]
\[=\binom{n-1}{n-1}+\binom{n-1}{n-2}+\dots+\binom{n-1}{n+1-i}+j.\]
Hence, using Claim 1 again, we find that the $k$-th term from the right in $R_n$ is none other than the $j$-th term from the right in $Q_{n(n+1-i)}$. By Claim 3, this implies the statement of the problem.[/hide]


For me, in the forum index, the high school forums are now mysteriously hidden. I double-checked my profile hide forums preferences, and they are set to be shown.

Hmmm they seem to be showing for me

Its probably a per-user thing

Erasing the cookies should solve the problem.

What I did was just reapply the preferences by pressing submit and it worked.

[quote="Temperal"]I have the same problem, but I had to delete the cookies related [b]and[/b] submit again in prefs. Is this related to the PHP parser?[/quote]Not sure why this happened. I could not reproduce the bug unfortunately.

Some lucky star shined and I was able to reproduce and diagnostic the bug. Should be fixed now :D


I'll try to buy them all.  However, because of the slim chance of that happening ill state my console prefrences:

1. Nintendo-revolution

2. PS3

3. X-box 360

neither my parents dont allow video games sux 4 me

i'm definitely going to get an x-box 360.  I have an x-box now and it has everything i could want.  i don't think that nintendo is going to last much longer in the game console businiess so i am not considering that. Sony is fine but the x-box 360 is coming out way before the ps3, so why wait?

i dont really know if I'll even choose to buy one... i dont have the time, but if i do it'll be between the revolution and PS3... i cant decide which one i'll like better..

[quote="yif man12"]there's another topic on the same exact thing in F&G.[/quote]

"another"? Where? :P

I'll probably buy the PS3 or the XBox 360.

Well, not exactly the same, but discusses PS3 and Xbox360.

http://www.artofproblemsolving.com/Forum/viewtopic.php?t=38226

Xbox-360. definitely. If they launch Halo3 with that, now they will dominate.

[quote="xxreddevilzxx"]If they launch Halo3 with that, now they will dominate.[/quote]

Halo3 is due out next spring. Apparently Microsoft is releasing it the same day PS3 comes out.

their probably trying to kill there competition

ps3 and xbox 360 both rockssssssssssssss
i don't know about the other one

[quote]MithsApprentice
PostPosted: Tue May 31, 2005 12:09 pm    Post subject:
Yes that's the more transparent facet of their marketing strategy. But I don't understand why they wouldn't release it a day before or something like th at. Doing it on the day of compromises their business as well.[/quote]
although it will comprimise some business some ppl will already have the xbox 360 and if they can get the halo3 which is very popular instead of buying a whole new gaming console plus games which would cost a lot of money they would get rid of the competition and for those ppl who like to wait for the prices to go down word of mouth goes around and ppl learn how great halo3 is and every1 buys that instead of the ps3 becuz there isnt a specific game that is as popular as halo3 so ppl forget about it

errr.. sorry havent thought of that :blush:

i'll be waiting for a price drop (or two) before i get the xbox 360.

[quote="G-UNIT"]whats f and g?[/quote]Actually is G&F right now :D He means this forum :)

looks like g-unit is the lone wolf :(  :(  :(  :(  :(

[quote="Super Naterul"][quote="NoSoupForYou"]I hate buying new consoles, because all the games I have will be incompatible.[/quote]

Actually I think the new consoles use old games.(revolution using gamecube, Ps2 using Ps3. xbox 360 using xbox)[/quote]

Really?

yes nintendo revolution i believe is gamecube games accesible, same with ps3 im not sure about x-box

All the systems will be backwards compatible. I'm going to get the Xbox 360 cuz i love the xbox i have now, and Halo dominates all. The PS3 I think though, is a more powerful system in specs, because of the cell processor. I still back Microsoft all the way. As to the Revolution... the Gamecube's developers didnt take advantage of the gamecube's power, so the graphics in almost all of the games werent amazing.. maybe gamecube was harder to work with? I probably wont buy the Revolution but i'll definitely take a look at it.

i like anything that's made by nintendo because they have a of exp. in making games.

I'm not going to get any of them...

PC ALL THE WAY!! YEAH!!!!

[quote="MithsApprentice"][quote="A+MATH"]PS3's cell processor and shaders are so much more advanced than anything the XBOX-360 has that Microsofts current marketing stategy is to sell each console at a loss of 240$ hoping that revenue from games will make up for the loss. I think Microsoft sux![/quote]

That's not a technique that's identified with Microsoft specifically. Almost every console is cheap enough so that the company selling it loses money on the console and then makes money off the games you buy.  I'm pretty sure Sony sells PS at a similar deficit.

As for noneoftheabove, that doesn't explain why Microsoft would release it on the same day as PS3. In fact, that's even more reason to give Halo a few days head start.[/quote]

Why can't sony just come back and push their release date up a day or 2 (if they were concerned about Halo 3)?

Also, xbox 360 will have perfect dark zero. :)   it'll probably be cooler than halo.

[quote="d343seven"]
Why can't sony just come back and push their release date up a day or 2 (if they were concerned about Halo 3)?[/quote]

Apparently Microsoft will announce Halo's release date after Sony announces theirs. And really, as far as marketing goes, it's stupid to change your release date after announcing (internationally btw) what it will be.

I remember when getting a Game Boy Color for Christmas would be considered the happiest moment of one's life. Now, we're so anxious for the companies to give us another console. Does anyone take time to actually enjoy the games?

For example, I currently own a Gameboy color, an SNES, and an N64. For GBC I have several old-school-revival games such as Marble madness, Super mario bros (NES), Ms. Pacman, etc. For SNES I have F-zero (that's all I can remember at the moment) and for N64 I have Tetrisphere, Supersmash bros, Perfect dark, and Mario Kart 64.

And Tetrisphere is my current favorite. I'm trying to beat the game. Which is fun and which takes up good amounts of time when I'm bored.

So that shows you how behind-the-times I am... but I encourage everyone who knows their video games to take a little time and dig through your old stuff, and try to find a remnant of games long past. Beat Super Mario 3 again, or if you're extremely bored, play a few rounds of Paperboy.

[/my2cents]


LaTeX
number theory proposed
number theory

The sequence un is defined by u1=1,u2=3, un=(n+1)un-1 - nun-2.
Which members of the sequence that are divisible by 11.
[b]un means u(sub)n sorry for not using Latex[/b]

$ u_n \equal{} (n \plus{} 1)u_{n \minus{} 1} \minus{} nu_{n \minus{} 2}$

$ \Longleftrightarrow u_n \minus{} u_{n \minus{} 1} \equal{} n(u_{n \minus{} 1} \minus{} u_{n \minus{} 2}) \equal{} n(n \minus{} 1)(u_{n \minus{} 2} \minus{} u_{n \minus{} 3})$

$ \equal{} \cdots \equal{}n(n \minus{} 1)(n \minus{} 2)\cdots 3(u_2 \minus{} u_1) \equal{} n!$

$ \therefore u_n \equal{} u_1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1} (u_{k \plus{} 1} \minus{} u_k)\ (n\geq 2)$

$ \equal{} 1 \plus{} \sum_{k \equal{} 1}^{n \minus{} 1} (k \plus{} 1)! \equal{} \boxed{\sum_{k \equal{} 1}^n k!\ (n\geq 1)}$.

there is an answer of this problem but i am not sure that is true. Here it is  all n except 1,2,3,5,6,7,9.


Putnam
logarithms
real analysis
real analysis unsolved

Prove or disprove that for the sequence $a_{n}$ defined in Putnam B6 question, $a_{n}-n^{k/(k+1)}(1+1/k)^{k/(k+1)}$ is bounded.

For the benefit of other readers, the Putnam problem is discussed here:
[url]http://www.artofproblemsolving.com/Forum/viewtopic.php?t=122460[/url].

Yes, I believe your sequence is bounded.  In fact, I believe it converges to 0.  Below is my argument.

Let $b_{n}= a_{n}^{(k+1)/k}$.  As I showed in the link above, $b_{n}$ satisfies the recurrence relation
\[b_{n+1}= b_{n}+\frac{k+1}{k}+O\left(\frac{1}{b_{n}}\right) \, . \]
By the Putnam problem, we know that $b_{n}$ is asymptotic to $\frac{k+1}{k}n$.  Thus we have
\[b_{n+1}= b_{n}+\frac{k+1}{k}+O\left(\frac{1}{n}\right) \, . \]
By summing, we get
\[b_{n}= \frac{k+1}{k}n+O(\log n) \, . \]
Going back to $a_{n}$, we get
\[a_{n}= \left(\frac{k+1}{k}n\right)^{k/(k+1)}+O\left( \frac{\log n}{n^{1/(k+1)}}\right) \, . \]
The error term approaches zero.

My argument in that thread was precisely a proof that this sequence is bounded.


MATHCOUNTS geometry calculus function trigonometry quadratics USAMTS

Hi.  I want to buy a graphing calculator from Texas Instruments but am confused on which one.  I am either going to buy the TI-84 Plus Silver Edition or the TI-89 Titanium Edition.  I'm going to be doing MathCounts, AMC and other various math competitions, so I don't know which one too choose.  I'm thinking of the 89, but it seems like its hard to use.  Then the 84 looks old.  Which one do you have, and which would you suggest?

Thankx.

-vistamath

OK. I have the 84 Plus Silver Edition. Personally, it works great--easy to use and learn. Doing competitions, you also have more time to study instead of trying to figure out how to use the calculator (compared to the 89). The 89 is definitely more powerful, though. Also, 84 is allowed on almost everything...well basically the 84 works the better for me. You can also get free program/application downloads from ticalc.org, and a few applications that help me a lot:

CabaMap
Inequal
Noshell
Omnicalc
PlySmlt2

Um...the 84 language is easier to learn (well, according to some code I've seen), and pretty decent.

So I would recommend 84. But then again, any calculator, as long as you can use it well.

I would recommend the 84, since the 89 is more of a Blackberry with it's functions.

The 84 Silver lags a lot, most likely due to it's large number of apps. I'd recommend deleting all the useless apps (which might take you a while) once you get it.

Yes, I did that--for example, all of the language apps.

Are there any people that would reccamend TI-89?  Also, would it help if I would say that I would be taking Honors Geometry this year and Algebra 2.  Which calculator would you use for those?

TI-89 definitely helps for calculus. It also has a computer algebra system that can simplify expressions with variables and whatnot for you. So I recommend TI-89.

My brother has a TI - 89 and i think its really good, but then again that's the only graphing calc i have used and the first time a saw a graph calc was 6 months ago (when my brother bought it, he's in college :D)

I've had an 89 since middle school, and I love it.  You get used to the interface and how to use it pretty quickly; at this point, I'm so used to using the 89 that I never want to go back and learn to use an 84.

As I said, it's all what you're used to. I just got the 84 because I didn't really know about anything when I got it and it was pretty popular at my school...so blablabla. Anyway, maybe I should get an 89 someday so I can do calculus and play Mario... :rotfl:  :rotfl:  :rotfl:

[quote="math154"] and play Mario... [/quote]

You can play Mario on the 83+ and 84+, as well, provided you have some sort of OS application, such as Mirage, and you are willing to devote about 75% of your memory to Mario alone.  :P

I know; but the graphics suck, and the Silver Edition has an OS conflict with the program itself, so the LCD thing is screwed up. btw, I already tried like all of the games. And I am a programming/calculator geek...

Crunchy OS totally owns Mirage (not really).

get the 89T. It's not hard to learn how to use, and it'll help you out in the long run.

Pretty print helps, and if you're going to mess with TI-BASIC on the calc, the 89 is much easier to code with than the Z80 (83, 84, etc) calculators.

AND it's allowed on the SAT :)

hmm.. I've had mine for 2.5 years.. got it at OfficeMax five minutes before it closed XD

Hm...well...I have a program that lets me go up to 24000 digits of calculation. I have a summation program. I have an equation solver program. I have an algebraic expression simplifier program. I have an exact trig value program. You see, the TI-89 IS better, but you don't really need to spend the extra money.

Plus there are like tons of cool nifty applications for the 84 that I haven't mentioned.

Hm...why am I defending the 84 when it's obvious the 89 is better... :rotfl:  :roll:  :rotfl:

anyway I'm just used to the 84...and z80 assembly code is PWNAGE.

I have a 89 and a 84 and personally I find the 84 much easier to use. I am not at a very high level of math, so I find the 84 better, but the 89 can definetly do a lot more. But when it comes to competetions, both will get the job done.

Anyways, I like the CAS of the 89, but if you're going to do local competitions in high school (assuming said competitions run in the same way as those in my state), I doubt the competitions organizers will let you use a symbol-manipulative calculator of any kind (that is to say, anything with a cas).

[quote="ernie"]IDK much about calcs, but how do the TI-89 and TI-84 rank with the TI-[i]n[/i]spire?[/quote]

TI-[i]n[/i]spire's are supposed to be better and are priced the same as the 84 and 89 (regular and CAS respectively), but I like the 84 and 89 better.

Oh if this is MathCounts, do not get TI-89. It's too good for MC. You can't use calculators for AMC. If you are doing USAMTS, use the computer instead.  TI-89 fails for programming basic formula races.

I recommend TI-Nspire or TI-84 for mathcounts.

what is a TI- Nspire.
 :maybe:  :maybe:  :maybe:
 I have a TI-84.

[quote="moogra"][quote="herefishyfishy1"]On the TI-89 titanium
You can program functions (Makes Pythagorean theorem/Heron's formula very easy to access)
You can do summations (an EXTREMELY useful feature)
You can solve equations (You will use this a lot in Mathcounts, trust me)
You can simplify algebraic expressions (You will sometimes need to do this)
You can get exact values of trigonometric functions (you might use this for a geometry problem)

On the TI-84 plus silver
You cannot program functions
You cannot do summation
You cannot solve equations
You cannot simplify algebraic expressions
You cannot get exact values of trigonometric functions

The TI-89 titanium has 188KB of RAM and 2.7 MB of flash ROM.
The TI-84 plus silver has 24 KB of RAM and a measly 480 KB of flash ROM.

Oh and in my opinion, the TI-89 programming language is loads easier to use than the TI-84 programming language.[/quote]

You have some of your facts wrong.
I can program any formula or summation using TI-84+.
Heres pythagorean theorem as it's short I'll post it.
:Prompt A,B
:Disp √(A²+B²
It is much harder on a TI-89. I know this for a fact. Quadratic formula and Herons are also 2 lines long (Prompt variables, disp the formula)

I can solve equations using strings on a TI-84+ (self programmed -- not some Mirage OS app) even though it is lame, it works. I can also graph 2 lines/equations and use calc->intersect. Who really needs exact values of trig functions?

TI-84+ Silver has 1.5MB of ROM.

I've used both 89 and 84+. The 84+ is definitely easier to use but if you want to devote some time into the 89, get it. I find reading the manual during boring times (airplane trips, car rides) entertaining when there is nothing better to do. I suggest 84+ for EASIER programming and 89 for more powerful calculations.[/quote]

OK, There's a difference between a [b]function[/b] and a [b]program.[/b] It is so easy on a TI-89 to do this:

hyp(x,y)
Func
√(x^2+y^2)
EndFunc

That's it. The Func and EndFunc are already written in for you, so you are really only writing one line of code. And you can call the function hyp(x,y), and instead of opening another program and asking for user input, it will simply compute it based on the argument.

Also: TI-89s give EXACT VALUES.

Use TI-89 Titanium for MathCounts!

[quote="AIME15"]Wait-what types of calculators are allowed at mathcounts?  Can you use calcs during the cd round?[/quote]

Any calculator that does not have a QWERTY keyboard is allowed. Calculators are used in all rounds except the Sprint and Countdown rounds.

[quote="AIME15"]So any calculator allowed on the SAT?[/quote]

Not all.  I think the TI-89's and TI-92's are not allowed.  Not 100% sure.

:blush: i didnt even know they had a TI92

Really? They have a TI-92?? I didn't know that...

[color=blue]* AIME15 and isabella2296 are humiliated.[/color]

........
spam
anyway, any graphing calc, (but a newer one is better) will help a lot
unfortunately, i only have a TI-34II

My bro has the TI-84+ Silver Edition and its perfect!  He jsut spends too much time on games.

I have a TI-N spire (with a TI-84 plus keypad for compatibility).  Like I have said in a post before, I use my Casio FX-115 ES too, scrappy calc that can do everything except graph.  :P

For a normal scientific calculator, I would recommend that Casio.

[quote="AIME15"]How much does that casio cost?

You won't need to graph much on mcs rit?[/quote]

Not to be a salesperson, but i think it costs either 10 or 15 bucks.  Good deal.

[quote="Math Geek"]Okay, no offense, but isabella2296 and AIME15 are treating the forum way too much like a chat room.  I would reply to any of the other numerous topics that have been spammed by the two, but they've all been locked.  Seriously, if you're just going to reply to your friend with a meaningless sentence or two that has nothing to do with the topic, then you don't need to use the forum, and no serious AoPS user needs you two to spam up every topic there is.

And, in case you two didn't know, admins can make the spam posts not count towards your overall post count if not completely ban you from AoPS. :wink:[/quote]

I tried addressing the issue on the round table.

Well at any rate, this thread has gone past its expiration date.

Mods please lock

I realize this post is spam, and I'm sorry.

@Math Geek:

I recently stopped spamming. I don't post for a higher post count. I sincerely didn't know they had a TI-92, though I realize that post was completely unnecessary. I apologize for the amounts of spam I have caused, but like I said, I stopped.


Prove that if $ a^{2}[b\plus{}c],b^{2}[a\plus{}c],c^{2}[a\plus{}b]$ are in a arithmetic progression then  $ a,b,c$ are in A.P

[hide="Hint"]
The distinct numbers $ a,\ b,\ c$ are in arithmetic progression if and only if $ \frac{b\minus{}a}{c\minus{}b}\in\mathbb{Q}$.
[/hide]

[hide="Solution"]
$ a^2(b\plus{}c),\ b^2(c\plus{}a),\ c^2(a\plus{}b)$ are in arithmetic progression $ \Longleftrightarrow \frac{b^2(c\plus{}a)\minus{}a^2(b\plus{}c)}{c^2(a\plus{}b)\minus{}b^2(c\plus{}a)}\in\mathbb{Q}$

$ \Longleftrightarrow \frac{b^2c\plus{}b^2a\minus{}a^2b\minus{}a^2c}{c^2a\plus{}c^2b\minus{}b^2c\minus{}b^2a}\equal{}$ $ \frac{c(b\minus{}a)(b\plus{}a)\plus{}ab(b\minus{}a)}{a(c\minus{}b)(c\plus{}b)\plus{}bc(c\minus{}b)}\equal{}\frac{(b\minus{}a)(bc\plus{}ac\plus{}ab)}{(c\minus{}b)(ac\plus{}ab\plus{}bc)}\equal{}\frac{b\minus{}a}{c\minus{}b}\in\mathbb{Q}\ \Longleftrightarrow a,\ b,\ c$ are in arithmetic progression.
[/hide]

[hide="One liner"]
Add $ abc$ to each term as that wont make any difference so we get
$ a(ab\plus{}bc\plus{}ac),b(ab\plus{}bc\plus{}ac),c(ab\plus{}bc\plus{}ac)$ and then divide by $ (ab\plus{}bc\plus{}ab)$..

Hence Proved

[/hide]

[quote="RAGHAV GROVER"][hide="One liner"]
Add $ abc$ to each term as that wont make any difference so we get
$ a(ab \plus{} bc \plus{} ac),b(ab \plus{} bc \plus{} ac),c(ab \plus{} bc \plus{} ac)$ and then divide by $ (ab \plus{} bc \plus{} ab)$..

Hence Proved

[/hide][/quote]
Very Nice Solution.Thanks :wink:

The question is actually false: for $ a \equal{} 1$, $ b \equal{} c \equal{} 0$ we have that $ 0, 0, 0$ are in arithmetic progression but $ 1, 0, 0$ are not.  Who wants to fix the question and the solutions?  ;)

Should it be such that $ ab \plus{} ac \plus{} bc \neq 0$?

Yes, that works fine.  A more natural condition is that the original progression be non-constant, but the purpose is exactly the same -- patching up a potential division by zero in both Mateescu's and Raghav's proofs.  (Incidentally, Mateescu solved a slightly different problem because of a linguistic misunderstanding: "are in arithmetic progression" means "are [b]consecutive[/b] terms in an arithmetic progression.")


Using the digits 1, 2, 3, 4, and 5, how many five-digit positive integers can be formed if only the digit 5 can be repeated any number of times?

With 2 5's, it should be 4c3=4 ways to pick the numbers multiplied by 5!/2=60 ways to rearrange, so actually its 240 for 2nd possibility. SImilarly for 3 its 6x5!/3!=120 for 4 its 4x5!/4!=20 and for 5 its still 1.

Thus we have 120+240+120+20+5=505


geometry
circumcircle
geometric transformation
geometry open

It is a well-known result that the pedal triangle of the first isodynamic point of triangle ABC is equilateral.  Is it true that in general this is the smallest equilateral triangle that can be inscribed in triangle ABC?  If so, is there a nice proof?

Take any equilateral triangle $DEF$ inscribed in $ABC$, with $D,E,F$ on $BC,CA,AB$ respectively, and let $P$ be the point where the circumcircles of $AFE,BDF,CED$ intersect (a simple angle chase will show that they do indeed intersect in a point). Now prove (again, an angle chase) that the pedal triangle of $P$ is also equilateral, so the smallest equilateral triangle inscribed in $ABC$ must be a pedal triangle, and there are only two points with equilateral pedal triangles, namely the isodynamic points.

Wow, that was quick.

Certainly the circumcircles of AFE, BDF, CED concur, namely at their Miquel point.  However, I don't follow the rest of the proof.  The pedal triangle of $P$, is it the same as $DEF$?  Why must the smallest equilateral triangle in $ABC$ be a pedal triangle?

It's not the same, it's just similar to it but smaller. This means that whenever you have an inscribed equilateral triangle, you can get a smaller one which is also pedal.

The pedal triangle of $P$ is obtained from $DEF$ through a spiral similarity centered at $P$ and through which $PD,PE,PF$ are turned into the perpendiculars dropped from $P$ to $BC,CA,AB$ respectively, so the distances decrease through this spiral similarity, meaning that the new triangle is smaller.

Ah, I see now.  I blame my slowness on the hot weather. :P  Very nice, thanks.


I'm having trouble w/ nested radicals...somebody help me please!

Simplify $ \sqrt{14 + 8\sqrt{3}}$.

I set it to $ a+b$ and got $ 14+8\sqrt{3}=a^2+2ab{}^b{}^2$, but I'm unclear on how to proceed here.

The general strategy is to set it equal to $ \sqrt a\plus{}\sqrt b$ (in problems like this) or $ \sqrt a\minus{}\sqrt b$ (in other problems like simplifying $ \sqrt{14\minus{}8\sqrt3}$).

To be more detailed:

[hide]We hopefully have that $ \sqrt {14 \plus{} 8\sqrt {3}} \equal{} \sqrt {(\sqrt {a} \plus{} \sqrt {b})^2}$, where

$ a \plus{} b \equal{} 14$ (integer part)
$ ab \equal{} 48$ ($ 8\sqrt {3} \equal{} 2\sqrt {48}$)

You can solve this in your head to get $ a \equal{} 6$, $ b \equal{} 8$ (or vice-versa), leading to

$ \sqrt {14 \plus{} 8\sqrt {3}} \equal{} \sqrt {(\sqrt {6} \plus{} \sqrt {8})^2} \equal{} \sqrt {6} \plus{} 2\sqrt {2}$[/hide]

Ok, so if i was solving $ \sqrt{49 \minus{} 20\sqrt{6}}$, i would do:

$ \sqrt{49\minus{}20\sqrt{6}}\equal{}\sqrt{(\sqrt{a}\minus{}\sqrt{b})^2}$, where $ a\minus{}b\equal{}49$ and $ ab\equal{}20$? And solve that?

That's correct. There's not always a solution, though.

Edit: Sorry, I just meant that AIME15's technique was correct--I didn't look at the numbers.

[quote="AIME15"]Ok, so if i was solving $ \sqrt {49 \minus{} 20\sqrt {6}}$, i would do:

$ \sqrt {49 \minus{} 20\sqrt {6}} \equal{} \sqrt {(\sqrt {a} \minus{} \sqrt {b})^2}$, where $ a \minus{} b \equal{} 49$ and $ ab \equal{} 20$? And solve that?[/quote]

No, that's not correct. Notice that $ (\sqrt a \minus{} \sqrt b)^2 \equal{} a \minus{} \mathbf{2\sqrt {ab}} \plus{} b$, so we have $ a\plus{}b \equal{} 49$ (not $ a\minus{}b\equal{}49$), but $ 2\sqrt {ab} \equal{} 20\sqrt 6 \implies \sqrt {ab} \equal{} 10\sqrt 6 \implies ab \equal{} 600$.

So if $ a \plus{} b \equal{} 49$ and $ ab \equal{} 600$, substitute $ b \equal{} 49\minus{}a$ to get $ a(49\minus{}a) \equal{} 600 \implies a^2\minus{}49a\plus{}600 \equal{} 0$, and solve for $ a$. In this case, it just happens that $ b^2\minus{}4ac\equal{}1$, so the answer is $ (49 \pm 1)/2 \equal{} 25\,\text{or}\,24$.

This corresponds to $ 5\minus{}2\sqrt 6$ and $ 2\sqrt 6\minus{}5$, which are the two square roots of the given expression.

[quote="Yongyi781"]
So if $ a \minus{} b \equal{} 49$ and $ ab \equal{} 600$, substitute $ b \equal{} a \minus{} 49$ to get $ a(a \minus{} 49) \equal{} 600 \implies a^2 \minus{} 49a \minus{} 600 \equal{} 0$, and solve for $ a$.[/quote]

But it's a sum ($ a \plus{} b \equal{} 49$), which leads to $ b \equal{} 49 \minus{} a$, and thus $ a(49\minus{}a) \equal{} 600 \implies a^2 \minus{} 49a \plus{} 600 \equal{} 0$

So that gets a=25 and b=24 (or vice versa) so $ \sqrt{49\minus{}20\sqrt6}\equal{}|5\minus{}2\sqrt6|\equal{}5\minus{}2\sqrt6$

Always make sure when its $ (a\minus{}b)^2$ that the greater one goes first (unless you are solving for a negative root).

OHHH, [i]this[/i] is how you do it. I've been trying to figure this out forever.

Umm... We didn't really need to know that.

Well, I decided to be polite and thank them for their help. I apologize if I have bored you with my mundane "thanks".

No, we're glad you learned something, but just think how hard it would be to read threads if the posting culture was to post every comment that ran through one's head! Before posting, ask yourself if you think it would contribute to the discussion.

An exception is thanking someone for answering your question, which is just a courtesy, but that too, if you keep doing it within the same thread, will get annoying. It's not that we're heartless, we just want to keep the threads readable.

By the way, welcome to AoPS :)

Another way to do t is to have this:

$  \\
\begin{eqnarray*} \\
a + b\sqrt {3} & = & \sqrt {14 + 8\sqrt {3}} \\
a^2 + 2ab\sqrt {3} + 3b^2 & = & 14 + 8\sqrt {3} \\
ab & = & 4 \\
a^2 + 3b^2 & = & 14 \\
\end{eqnarray*} \\
$

We find $ a$ and $ b$ and see it is $ 2\sqrt {2}$ and $ \sqrt {2}$ respectively.  Thus the answer is $ \boxed{2\sqrt {2} + \sqrt {6}}$

However, note that it is easier (generally) to equate as $ \sqrt a\plus{}\sqrt b$ or $ \sqrt a\minus{}\sqrt b$, since if there is a simplification, $ a,b$ must both be integers.

Actually, you may revive topics if you have something to add or to ask. Then again, that applies mostly in competition threads, where it would be better to post in the thread so other users who don't understand something can see it.

In this case, it probably would've been better to post a new thread, but it is related, so bleh.

Have you tried using the above method on this problem? You learn more if you actually try. At first glance, the problem doesn't appear to simplify nicely.


right now, i can't decide b/w Jan. test and March test.....
(i'm a junior)...
I got a pretty low score on PSAT (probably the lowest among people in this forum) and really confused when to take it...
i'd really appreciate any suggestions....

One more question: what do you guys think about Collegeboard Online thing? (i've already done the studyguide)

Thanks,

Do you feel you are ready for the January test ?
if not, you probably need more time to study


Does anyone know of any good, free Computer Algebra Systems?  I am currently using yacas, but it is too difficult to use, lacks many features, deoesn't work well on windows, etc.

I was extremely bored one day, so I wandered around Sourceforge for an hour and found [url=http://maxima.sourceforge.net/]Maxima[/url]


10 men can build 2 boats in 4 days. Hom many men are needed to build 5
boats in 2 days

Is there a formula that can always be used in these three variable questions?

These questions aren't about formulas as much as they are about common sense.  The common-sense propositions are these:

- If you have more men, you can build more boats.  Or, if you have more men, you can take less time to build the same number of boats.
- If you have more time, you can build more boats.  Or, if you have more time, you can use less men to build the same number of boats.
- If you have more boats, you need more men.  Or, if you have more boats, you need more time to build them.

Together these propositions imply that

$\frac{ \text{Boats}}{ \text{Men}\times \text{Time}}$

Is a constant (in fact, it's a measure of their productivity, which you simply then assume is constant).  If you want to approach such questions formulaically, you can using that, but I would rather do them [hide="like this"] $10$ men can build $2$ boats in $4$ days, so they can build $1$ boat in $2$ days.  If we want five times as many boats, we need five time as many men; hence, $\boxed{50}$ men. [/hide]

Thats good to know, so instead of looking for formulas I can just focus on the logic. Thank you


If $\dfrac{1}{a+ \omega } + \dfrac{1}{b+\omega} + \dfrac{1}{ c + \omega} + \dfrac{1}{ d + \omega } = \dfrac{2}{\omega}$ , $a,b,c,d \in \mathbb{R}$ , $\omega$ is the nonreal cube root of unity, prove that ... $a+ b+c+ d = 2abcd$

I use "w" instead of omega.  We have $w^2 + w + 1 = 0$.

Multiply out.  We get $\sum\limits_{a,b,c,d} w(a+w)(b+w)(c+w) = 2\prod (a+w)$

The left side is
$\sum w( w^3 + w^2(a+b+c) + w(ab+bc+ca) + abc)$

$= 4w + 3(a+b+c+d) + 2(ab+ac+ad+bc+bd+cd)w^2 + w(abc+abd+acd+bcd)$

The rightside is $2 (w^4 + (a+b+c+d)w^3 + (ab+ac+ad+bc+bd+cd)w^2 + (abc+abd+acd+bcd)w + abcd)$

Equating, we get left over $2w + (a+b+c+d) = (abc+abd+acd+bcd)w + 2abcd$

Equating real and imaginary parts, $a+b+c+d = 2abcd$.