sum of digits
number theory
Digits
Divisibility
IMO Shortlist

Prove that for each positive integer $n$, there exists a positive integer with the following properties: It has exactly $n$ digits. None of the digits is 0. It is divisible by the sum of its digits.

Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: [url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]introduction for the IMO ShortList/LongList project[/url] and regarding[url=http://www.mathlinks.ro/Forum/viewtopic.php?t=15623]solutions[/url]  :)

We prove by induction.
First we prove for $ 3^k$ digits. In this case we claim $ 3^k$  $ 1$s will satisfy the sequence.
$ 111 \equal{} 37*3$, so it is true for $ k \equal{} 1$. Assume it is true for $ k$. Then $ 3^k|\frac {10^{3^k} \minus{} 1}{9}$. So we need to prove $ 3^{k \plus{} 1}|\frac {10^{3^{k \plus{} 1}} \minus{} 1}{9} \equal{} (\frac {10^{3^k} \minus{} 1}{9})(10^0 \plus{} 10^k \plus{} 10^{2k})$. We have $ 3^k$ divides what is in the first parentheses, and $ 3$ divides what is in the second, since the sum of the digits of it is $ 3$. so $ 3^{k \plus{} 1}$ divides this and we are done.

Now we subtract a number $ m$ from $ 3^k$ $ 1$s such that the sum of the digits is still $ 3^k$, but there is one less digit, and the number is still divisible by the sum of its digits. To do this, first remove the first $ 1$, i.e. subtract $ 10^{3^{k \minus{} 1}}$. Now we need to add $ 10^a$ such that $ 3^k|10^a \minus{} 10^{3^{k \minus{} 1}} \equal{} 10^a(1 \minus{} 10^{3^{k \minus{} 1} \minus{} a})$ which is equivalent to $ 3^k|(10^{3^{k \minus{} 1} \minus{} a} \minus{} 1)$, which is just $ 3^{k \minus{} 1} \minus{} a$ $ 9$s. Since $ 3^k|\frac {10^{3^k} \minus{} 1}{9}$, $ 3^k|10^{3^{k \minus{} 2}} \minus{} 1$. So let $ 3^{k \minus{} 1} \minus{} a \equal{} 3^{k \minus{} 2}$, i.e. $ a \equal{} 3^{k \minus{} 1} \minus{} 3^{k \minus{} 2}$. So we just subtract $ m$ which will cause a $ 1$ further on to increase to $ 2$, reducing the number of the digits by $ 1$ but retaining the divisibility by the sum of the digits. If the first digit is $ 2$, we subtract $ m$ twice, if it is $ 3$ we subtract $ m$ $ 3$ times which will cause a $ 1$ further on to increase to $ 4$ and so on.

We will never be forced to add $ 1$ to a digit which is already $ 9$, since we can terminate the process when all digits are $ 9$, which is when there are $ 3^{k \minus{} 1}$ digits.
This is because $ \frac {3^k}{3^k \minus{} 1 \minus{} (3^k \minus{} 3^{k \minus{} 2} \minus{} 1)} \equal{} 9$, so when each time we remove the front digit we add to the digit $ 1/9$th of the length of the original number along from the digit we subtracted, meaning we will end with $ 3^{k \minus{} 1}$ $ 9$s.

So since this is true for all numbers of the form $ 3^k$, and all numbers between $ 3^k$ and $ 3^{k \plus{} 1}$ it is true for all $ n$

I'm confused about this solution. Can you apply the process to 111111111?

I'll rewrite the proof since the version above has a number of typos which cause it to be very confusing.

We prove by induction.
First we prove for $ 3^k$ digits. In this case we claim $ 3^k$  $ 1$s will satisfy the sequence.
$ 111 \equal{} 37*3$, so it is true for $ k \equal{} 1$. Assume it is true for $ k$. Then $ 3^k|\frac {10^{3^k} \minus{} 1}{9}$. So we need to prove $ 3^{k \plus{} 1}|\frac {10^{3^{k \plus{} 1}} \minus{} 1}{9} \equal{} (\frac {10^{3^k} \minus{} 1}{9})(10^0 \plus{} 10^k \plus{} 10^{2k})$. We have $ 3^k$ divides what is in the first parentheses, and $ 3$ divides what is in the second, since the sum of the digits of it is $ 3$. so $ 3^{k \plus{} 1}$ divides this and we are done.

Now consider the $ q$ digit long number between $ \frac {10^{3^k} \minus{} 1}{9}$ and $ \frac {10^{3^{k \minus{} 1}} \minus{} 1}{9}$ which works. We construct a number of $ q \minus{} 1$ digits which works. We claim subtracting a multiple of $ m \equal{} 10^{3^k \minus{} r} \minus{} 10^{3^k \minus{} r \minus{} 3^{k \minus{} 2}}$ preserves both the sum of the digits and the property that the number is divisible by its digits, where $ r\minus{}1$ is the number of digits less than $ 3^k$ we have.

Indeed, it is clear that subtracting $ m$ (or a multiple of $ m$) preserves the sum of the digits, since it increases one digit by one and decreases another by $ 1$.

Also, $ 3^k$ divides our number to start with by the inductive hypothesis, and
$ 3^k|10^{3^k \minus{} r} \minus{} 10^{3^k \minus{} r \minus{} 3^{k \minus{} 2}}$ is equivalent to
$ 3^k|10^{3^{k \minus{} 2}} \minus{} 1$
But this is clear since we established above that $ 3^k|\frac {10^{3^k} \minus{} 1}{9}$./

We will never be forced to add $ 1$ to a digit which is already $ 9$, since we can terminate the process when all digits are $ 9$, which is when there are $ 3^{k \minus{} 1}$ digits.
This is because $ \frac {3^k}{3^k \minus{} r \minus{} (3^k \minus{} 3^{k \minus{} 2} \minus{} r)} \equal{} 9$, so when each time we remove the front digit we add to the digit $ 1/9$th of the length of the original number along from the digit we subtracted, meaning we will end with $ 3^{k \minus{} 2}$ $ 9$s.

So our construction is essentially as follows: look at a number of $ q$ digits with first digit $ s$ which works. subtract $ m$ $ s$ times. We now have a number of $ q \minus{} 1$ digits which works. (This fails for $ k \equal{} 1$ and $ k \equal{} 0$ since we use $ 3^{k \minus{} 2}$, but it is trivial to contruct examples for these cases)

For the $ 11111111$ example we have
$ 111111111$
$ 21111111$
$ 3111111$
$ 411111$
$ 51111$ (e.g. here $ r\equal{}5$, $ k\equal{}2$)
$ 6111$
$ 711$
$ 81$
$ 9$ (although we can stop when we reach $ 6111$ since $ 3$ digits is already a power of $ 3$)

Wow,very nice solution!Thank you so much!
Here are two solutions(the second is the official one) :lol:

Suppose $n$ is sufficiently large such that $n \geq 9(2+\lceil \log_2{n} \rceil)$( roughly something like $n \geq 81$ will work, though this can be optimised to be lower).
First of all, consider a number with $(2+ \lceil \log_2(n)) \rceil)$( we will call this number $k_n$ from now on) digits, such that none are zero, such that this number is a multiple of $2^{k_n}$( it is obvious that such a number exists, we can for instance make one which has digits $1$ and $2$ only).Because $n$ is sufficiently large, one can obviously make an $n$ digit number which has all nonzero digits, whose digit sum is $2^{k_n}$ and whose last $k_n$ digits are the same as that of the number we chose. Clearly, this number works. So we are finished for the problem if $n \geq 81$. The rest can be made through examples. For completeness, here are the examples for $n=1$ to $80$( which also shows a way for a second construction):
$n=1$ to $10$:
$1,12,132,1212,22225,222222,1111112,33333325,333333333,3333311125$
$n=11$ to $20$:
$33322211125,333221111125,3332111111125,33311111111125$ and this pattern(digit sum $25$, last $2$ digits $25$) can be followed up to $n=20$.
$n=21$ to $70$
Here we can make numbers with last two digits $25$ and digit sum $75$. This pattern can take us up to $70$.
$71$ to $80$:
Make numbers with last $3$ digits $125$ and digit sum $375$.
Using this strategy and extending it can be a second way to find a construction.
Proved.

The base cases are easy. Indeed, it is possible to see that $1$, $12$, $112$, $4112$ work.
For the next numbers, we will define a infinite sequence $(a_1,a_2,a_3,...)$.
Let $a_1= 1136$. We define $a_n = a_{n-1}+10^{n+2}c$, where we pick $c=1$ if $a_{n-1} \equiv 2^{n+2} $  (mod $2^{n+3}$) and $c=2$ if $a_{n-1}. \equiv  0 $ (mod $2^{n+3}$  ).
Hence, $2^{n+3} \mid a_n$ for all $n$.
It is well-known that $2^k \mid x$ if and only if the last $k$ digits of $x$ are divisible by $2^k$.
So we will construct a number with $a_n$ as it`s last $n+3$ digits and we will make the sum of all digits to be $2^{n+3}$
Observe that if $x=(11...11(a_n))$ and  $s(t)$ be the sum of digits of the number $t$, hence the largest number of digits we can make with $a_n$ is
$2^{n+3}- s(a_n)$
So we only have to verify if $$ 2^{n+3}-s(a_n)	+(n+3) \geq \lfloor \frac{2^{n+4}-s(a_{n+1})}{9}	\rfloor +(n+4)$$
That is, if we can make larger and larger numbers picking $a_{n+1}$. But this is inequality is obvious, since $s(a_{n+1})$ grows at most $2$ at time. :lol:

For $n=1$ use $1$, for $n=2$ use $12$, for $n=3$ use $112$. Let the current number be $k_n$, so $k_3=112$. We will separate the numbers starting from $k_3$ into the [i]free part[/i] $a_n$ and the [i]rigid part[/i] $b_n$. For $n=3$, let $a_n=1$ and $b_n=12$. We can get $k_n$ by concatenating $a_n$ and $b_n$. We will use an algorithm to generate $a_{n+1}$ and $b_{n+1}$ from $a_n$ and $b_n$.

- If $a_n$ is not a string of $1$s, subtract $1$ from a digit that is not yet $1$ and put a $1$ at the beginning of $a_n$ to get $a_{n+1}$, and let $b_{n+1}=b_n$.
- Otherwise, let the length of $b_n$ be $t$. Putting a $1$ or a $2$ at the beginning of $b_n$ will result in it having a $\nu_2$ of $t$ or at least $t+1$, so choose the one that gets it at least $t+1$, and that's our new $b_{n+1}$. Multiply $a_n$ by $9$ then keep docking off $1$s off non-$1$s(starting from the least significant digit first, say) until the sum of its digits plus the sum of the digits of $b_{n+1}$ is exactly $2^{t+1}$, and that's our $a_{n+1}$.

For the sake of clarity, here are the first few terms of $a_i$, $b_i$, and $k_i$:

\begin{tabular}{l|l|l|l}
$n$ & $a_n$ & $b_n$ & $k_n$ \\ \hline
$3$ & $1$ & $12$ & $112$ \\ \hline
$4$ & $4$ & $112$ & $4112$ \\ \hline
$5$ & $13$ & $112$ & $13112$ \\ \hline
$6$ & $112$ & $112$ & $112112$ \\ \hline
$7$ & $1111$ & $112$ & $1111112$ \\ \hline
$8$ & $7111$ & $2112$ & $71112112$ \\ \hline
$9$ & $16111$ & $2112$ & $161112112$ \\
\end{tabular}

After solving the problem, I wrote code to test that it works and to hopefully make my solution clearer.

[pywindow]
def helper(n):
	if n <= 2: raise Exception("pet the catloe")
	if n == 3: return [1,12]
	a,b = helper(n-1)

	firstCase = False
	for i in range(len(str(a))):
		if str(a)[i] != "1":
			firstCase = True
			a = list(str(a))
			a[i] = str(int(a[i])-1)
			a = "".join(a)
			a = "1" + a
			a = int(a)
			break

	if firstCase: return [a,b]

	# We are in the second case
	b1 = int("1"+str(b))
	b2 = int("2"+str(b))

	newB = ""

	if b1 % (2**len(str(b1))) == 0: newB = b1
	else: newB = b2

	newB = str(newB)

	a = "9"*len(str(a))

	while sum(int(x) for x in a) + sum(int(x) for x in newB) > 2**len(newB):
		for i in range(len(a)-1,-1,-1):
			if a[i] == "1": continue
			a = list(a)
			a[i] = str(int(a[i])-1)
			a = "".join(a)
			break

	return [int(a), int(newB)]

def solve(n):
	if n == 1: return 1
	if n == 2: return 12
	a,b = helper(n)
	return int(str(a)+str(b))

for n in range(1,31):
	answer = solve(n)
	if answer % sum(int(x) for x in str(answer)) == 0:
		print(answer,"is valid for n =",n)
	else:
		print(answer,"is INVALID for n =",n)
[/pywindow]

Lemma:
For any n and any s, such that n<=s<=9*n, there exists a n-digit number with digit sum s.
Proof: This is trivial: Use 11111…111 with n digits, and keeping adding one to a non-9 digit you pass through s by IVT on the way to 999…999.

Fix n, the number of digits. Let k=floor(log2(5*n)).
We will show that there exists a number of k digits, divisible by 2^k, which does not contain any zeros. Suppose the decimal representation of k contains t digits. Start with the string s=0000…000[2^k], k-t zeros at the front to make n digits in total. Here [2^k] is the decimal representation of 2^k. For each zero, starting from the right, we greedily make a new string s’, with that zero having been plugged in with some other digit, and no more zeros created to the right. At a certain step, suppose the latest zero in the number is in the 10^r place (as in, units place tens place hundreds place etc). Now add 10^r * 2^k to the number. This ends in r zeros, so no digit after is affected. The new number s’ is also divisible by 2^k iff s was, as we only ever added multiples of 2^k.
We keep doing this until the last n digits of the new s are 0, which will happen eventually as s has only finitely many digits. At this time s will have at most k+t+1 digits. This is because we never modified digits before the k+t+1 th from the end.

(the real point of the proof is that the effect on # of digits and sum of digits for large n of the last k digits is negligible)
Now consider the new s’, let it be o. Suppose o has p digits, k<=p<=k+t+1, and has a digit sum of S. We want the total digit sum to be 2^k, with n digits. We need to prove that (n-p)<=2^k-S<=9(n-p).
S<=(k+t+1)9 <= 18k. (as t<k)
We bound on 2^k.
2.5n<=2^k<=5n
2.5n-18k<=2^k-S<=5n-18k

we just must prove that (n-p) <= 2.5n-18k (inequality 1)
and that 5n-18k<=9(n-p) (inequality 2)
then we would be done by the lemma.

To prove inequality 1,
n-p <= 2.5n-18k
-p<=1.5n-18k
If 1.5n>=18*floor(log2(5n)), this is trivially true.

5n-18k<=9n-9p (inequality 2)
-18k <= 4n-9p
well here, p<=2k as k+t+1<2k
so RHS >= 4n-18k >= LHS as n is positive.

So inequality 2 is always true. Now for inequality 1:
1.5n>=18*floor(log2(5n))
n >= 12*floor(log2(5n))
n >= 12*floor(log2(n))+36
If n>108 this is trivial to prove using induction. So the proof is complete with some small case checking (only 108 cases).
These are solutions for 9 to 108, generated with this python program (with 1 to 8 added without the program)

[pywindow]from math import *
from copy import *
def genersum(numdig, sumdig):
    orsumdig=copy(sumdig)
    ornumdig=copy(numdig)
    #generate number with numdig digits, summing to sumdig
    number=''
    for i in range(copy(numdig)):
        numdig-=1
        number+=str(min(9,sumdig-numdig))
        sumdig-=int(number[-1])
    if sumdig<0:
        return '-1'
    if sum(map(int,number))!=orsumdig or len(number)!=ornumdig:
        #print(sum(map(int,number)),len(number))
        return '-1' #impossible
    return number
n=124 #number of digits
k=9 #width
def solve(n):
    #n is the number of digits
    k=floor(log2(5*n))
    u=2**k
    t=floor(log10(u))+1
    s=str(u)
    cur=u
    st=str(cur).zfill(k)

    while 1:
        if '0' not in st[-k:]:
            break
        tau=k-st.rfind('0')
        cur+=10**tau * u
        st=str(cur).zfill(k)

    su=sum(map(int,str(cur)))
    remsu=u-su

    remd=n-len(str(cur))
    rest=genersum(remd,remsu)
    if rest=='-1':
        return -1
    ans=rest+str(cur)
    ch1=(int(ans)%sum(map(int,ans)))==0
    ch2=(len(str(ans)))==n
    ch3='0' not in ans
    if ch1 and ch2 and ch3:
        return ans
    return -1
for i in range(1,1000):
    print(i,solve(i))[/pywindow]

[hide=Solutions for 1 through 108]
1 1
2 22
3 333
4 4444
5 55555
6 666666
7 7777777
8 88888888
9 992131232
10 9911131232
11 98111131232
12 971111131232
13 9951117494464
14 99411117494464
15 993111117494464
16 9921111117494464
17 99111111117494464
18 981111111117494464
19 9711111111117494464
20 96111111111117494464
21 951111111111117494464
22 9411111111111117494464
23 93111111111111117494464
24 921111111111111117494464
25 9111111111111111117494464
26 99999999811111111138368128
27 999999997111111111138368128
28 9999999961111111111138368128
29 99999999511111111111138368128
30 999999994111111111111138368128
31 9999999931111111111111138368128
32 99999999211111111111111138368128
33 999999991111111111111111138368128
34 9999999811111111111111111138368128
35 99999997111111111111111111138368128
36 999999961111111111111111111138368128
37 9999999511111111111111111111138368128
38 99999994111111111111111111111138368128
39 999999931111111111111111111111138368128
40 9999999211111111111111111111111138368128
41 99999991111111111111111111111111138368128
42 999999811111111111111111111111111138368128
43 9999997111111111111111111111111111138368128
44 99999961111111111111111111111111111138368128
45 999999511111111111111111111111111111138368128
46 9999994111111111111111111111111111111138368128
47 99999931111111111111111111111111111111138368128
48 999999211111111111111111111111111111111138368128
49 9999991111111111111111111111111111111111138368128
50 99999811111111111111111111111111111111111138368128
51 999997111111111111111111111111111111111111138368128
52 9999999999999999999991111111111111111111112662656256
53 99999999999999999999811111111111111111111112662656256
54 999999999999999999997111111111111111111111112662656256
55 9999999999999999999961111111111111111111111112662656256
56 99999999999999999999511111111111111111111111112662656256
57 999999999999999999994111111111111111111111111112662656256
58 9999999999999999999931111111111111111111111111112662656256
59 99999999999999999999211111111111111111111111111112662656256
60 999999999999999999991111111111111111111111111111112662656256
61 9999999999999999999811111111111111111111111111111112662656256
62 99999999999999999997111111111111111111111111111111112662656256
63 999999999999999999961111111111111111111111111111111112662656256
64 9999999999999999999511111111111111111111111111111111112662656256
65 99999999999999999994111111111111111111111111111111111112662656256
66 999999999999999999931111111111111111111111111111111111112662656256
67 9999999999999999999211111111111111111111111111111111111112662656256
68 99999999999999999991111111111111111111111111111111111111112662656256
69 999999999999999999811111111111111111111111111111111111111112662656256
70 9999999999999999997111111111111111111111111111111111111111112662656256
71 99999999999999999961111111111111111111111111111111111111111112662656256
72 999999999999999999511111111111111111111111111111111111111111112662656256
73 9999999999999999994111111111111111111111111111111111111111111112662656256
74 99999999999999999931111111111111111111111111111111111111111111112662656256
75 999999999999999999211111111111111111111111111111111111111111111112662656256
76 9999999999999999991111111111111111111111111111111111111111111111112662656256
77 99999999999999999811111111111111111111111111111111111111111111111112662656256
78 999999999999999997111111111111111111111111111111111111111111111111112662656256
79 9999999999999999961111111111111111111111111111111111111111111111111112662656256
80 99999999999999999511111111111111111111111111111111111111111111111111112662656256
81 999999999999999994111111111111111111111111111111111111111111111111111112662656256
82 9999999999999999931111111111111111111111111111111111111111111111111111112662656256
83 99999999999999999211111111111111111111111111111111111111111111111111111112662656256
84 999999999999999991111111111111111111111111111111111111111111111111111111112662656256
85 9999999999999999811111111111111111111111111111111111111111111111111111111112662656256
86 99999999999999997111111111111111111111111111111111111111111111111111111111112662656256
87 999999999999999961111111111111111111111111111111111111111111111111111111111112662656256
88 9999999999999999511111111111111111111111111111111111111111111111111111111111112662656256
89 99999999999999994111111111111111111111111111111111111111111111111111111111111112662656256
90 999999999999999931111111111111111111111111111111111111111111111111111111111111112662656256
91 9999999999999999211111111111111111111111111111111111111111111111111111111111111112662656256
92 99999999999999991111111111111111111111111111111111111111111111111111111111111111112662656256
93 999999999999999811111111111111111111111111111111111111111111111111111111111111111112662656256
94 9999999999999997111111111111111111111111111111111111111111111111111111111111111111112662656256
95 99999999999999961111111111111111111111111111111111111111111111111111111111111111111112662656256
96 999999999999999511111111111111111111111111111111111111111111111111111111111111111111112662656256
97 9999999999999994111111111111111111111111111111111111111111111111111111111111111111111112662656256
98 99999999999999931111111111111111111111111111111111111111111111111111111111111111111111112662656256
99 999999999999999211111111111111111111111111111111111111111111111111111111111111111111111112662656256
100 9999999999999991111111111111111111111111111111111111111111111111111111111111111111111111112662656256
101 99999999999999811111111111111111111111111111111111111111111111111111111111111111111111111112662656256
102 999999999999997111111111111111111111111111111111111111111111111111111111111111111111111111112662656256
103 9999999999999999999999999999999999999999999999993111111111111111111111111111111111111111111152255232512
104 99999999999999999999999999999999999999999999999921111111111111111111111111111111111111111111152255232512
105 999999999999999999999999999999999999999999999999111111111111111111111111111111111111111111111152255232512
106 9999999999999999999999999999999999999999999999981111111111111111111111111111111111111111111111152255232512
107 99999999999999999999999999999999999999999999999711111111111111111111111111111111111111111111111152255232512
108 999999999999999999999999999999999999999999999996111111111111111111111111111111111111111111111111152255232512
[/hide]

By LTE, $\nu_3(\tfrac19(10^n-1))=\nu_3(n)$, so $3^k\mid \underbrace{111\dots 1}_{3^k \text{ 1s}}$.

We also know that by LTE, $10^{2\cdot 3^{n-1}}\equiv 10^{3^{n-1}}\equiv 1\pmod {3^{n}}$ for all $n\ge 2$. We can now proceed with the proof for non-powers of $3$.

Let $3^k$ be the smallest power of three larger than $n$. Suppose $2\cdot 3^{k-1}\le n\le 3^{k}-1$. We begin with the number
\[\frac{1}{9}(10^n-1)=\sum_{i=0}^{3^k-1}{10^i}\]
We will modify this number while keeping the sum of digits equal to $3^k$ and also keeping the number itself divisible by $3^k$. Consider the number $10^{n+i}-10^{n+i-2\cdot 3^{k-1}}$ for some nonnegative integer $i\le 3^k-n-1$. Since $3^k\mid 10^{2\cdot 3^{k-1}}-1$, we have $3^k\mid 10^{n+i}-10^{n+i-2\cdot 3^{k-1}}$ by multiplying by $10^{n+i-2\cdot 3^{k-1}}$, which is an integer because of our definitions. Indeed, we have that the following number is divisible by $3^k$:
\begin{align*}
\sum_{i=0}^{3^k-1}{10^i}-\sum_{i=0}^{3^k-n-1}{10^{n+i}-10^{n+i-2\cdot 3^{k-1}}} &= \sum_{i=0}^{3^k-1}{10^i} - \sum_{i=n}^{3^k-1}{10^i} + \sum_{i=n-2\cdot 3^{k-1}}^{3^{k-1}-1}{10^i} \\
&= \underbrace{111\dots 1}_{n \text{ 1s}} + \underbrace{111\dots 111}_{3^k-n \text{ 1s}}\underbrace{000\dots 0}_{n-2\cdot 3^{k-1} \text{ 0s}}
\end{align*}
Which clearly has $n$ digits, as required. For example, when $n=2\cdot 3^{k-1}$, this number is
\[\underbrace{111\dots 1}_{3^{k-1} \text{ 1s}}\underbrace{222\dots 2}_{3^{k-1} \text{ 2s}}\]
Note that we are just moving a number of contiguous ones right $2\cdot 3^{k-1}$ place values to make the new number have $n$ digits. When $3^{k-1}\le n\le 2\cdot 3^{k-1}-1$, we can do the same thing but moving right $3^{k-1}$ place values. Since the writeup process is tedious and it is isomorphic to the previous case, I will not write it all out. Have an example instead:
\begin{eqnarray*}
f(9)&=&111111111 \\
f(8)&=&11111211 \\
f(7)&=&1111221\\
f(6)&=&111222\\
f(5)&=&11322\\
f(4)&=&1332\\
\end{eqnarray*}


Find all pairs of integers $(x, y)$ satisfying the equation \[y^{2}(x^{2}+y^{2}-2xy-x-y) = (x+y)^{2}(x-y)\]

bump...

What is the exact name of the contest? I tried googling to look up the solution, but it didn't come up with any math contest.


For the first one:
You want to find the value of $(x+y)^{2}$, which equals $x^{2}+2xy+y^{2}$.  Since you know $x^{2}+y^{2}$ and $xy$, you can find that value by adding.
$x^{2}+y^{2}+xy+xy=73+24+24$, so $x^{2}+2xy+y^{2}=121$, and that is the answer.

OK, a few exponent rules first.
\[\displaystyle a^{\frac{1}{2}} = \sqrt{a}\]
and
\[\displaystyle a^{-x} = \frac{1}{a^x}\]

Do you have enough to solve the question now? :)

[quote="laptof89"]I always have trouble manipulating exponents in algebric equations but ouce someone solves the problem for me it seems so easy. I think I just don't understand exponents very well. Anyway here are some problems taken from an SAT study guide book:

If x^2  + y^2 = 73 and xy = 24, what is the value of (x + y)^2 ?

I did solve this problem but by using trial and error by finding factors of 24 and than plugging them into the equation. However, I rather solve the problem by using actual methods that save time.

If (a + b)^1 / 2 = (a - b) ^ -1 / 2, which of the following must be true?

a. b = 0
b. a + b = 1
c. a - b = 1
d. a^2 + b^2 = 1
a^2 - b^2 = 1

Thank you for the help.[/quote]

[hide="Solution"]
(x + y)^2 = x^2 + y^2 = 2xy.
= 73 + 2(24)
= 73 + 48
= [b]121[/b]
[/hide]

Ok, for the first one I knew immidietly ounce I saw factoring. For the second one I got up to here:

sqrt(a + b) * sqrt(a - b) = 1 but not sure what to do from here.

Nevermind, I figured it out: a ^ 2 - b ^ 2 = 1. You guys make things so easy, too bad I can't take one of you with me this Saturday

[quote="laptof89"]I always have trouble manipulating exponents in algebric equations but ouce someone solves the problem for me it seems so easy. I think I just don't understand exponents very well. Anyway here are some problems taken from an SAT study guide book:

If x^2  + y^2 = 73 and xy = 24, what is the value of (x + y)^2 ?

I did solve this problem but by using trial and error by finding factors of 24 and than plugging them into the equation. However, I rather solve the problem by using actual methods that save time.

If (a + b)^1 / 2 = (a - b) ^ -1 / 2, which of the following must be true?

a. b = 0
b. a + b = 1
c. a - b = 1
d. a^2 + b^2 = 1
a^2 - b^2 = 1

Thank you for the help.[/quote]

[hide="solution"]
er...i think im confused, for the second part, does d. contain both $a^2 + b^2 = 1$ $and$  $a^2 - b^2 = 1$, or does it contain only $a^2 + b^2 = 1$ ??? if it contains both, then i choose d. but if it contains only $a^2 + b^2 = 1$ , then i choose $a^2 - b^2 = 1$...
now solution time, you could say that $\displaystyle\ (a + b)^{1/2} = \sqrt(a+b)\ $ that means that $\displaystyle\ (a-b)^{-1/2}= \frac 1\sqrt(a-b)\\ $  square each side, $\displaystyle\ a+b =\frac 1 {a-b} $ then cross multiply, $ (a+b)(a-b) =1 $ so that means $ {a^2}-{b^2}=1 $
[/hide]


search
number theory
prime numbers
algebra proposed
algebra

It is an open problem to show that there are infinitely many $ p,q$ prime with difference $ 2a$ for any fixed integer $ a$  :wink:

not enough

who can find a detailled solution

What's not enough??
If someone proves your problem he has proved that there exist infinitely primes with difference 6.
I'm saying that no one has proved this till now OK? (Ever heard of the twin prime conjecture for example?)
Of course I may be wrong... :roll:

[quote="greatestmaths"]prouve that there is an infinite natural relative  number p such that:

1) p, p+6 and p+12 are primes.
2)p+1,p+2,p+3,p+4,p+5,p+7,p+8,p+9,p+10 and p+11 aren't primes.[/quote]
and also about the source...  :huh:
I'm not sure this is from Iran,could you at least name the competition which this problem was proposed in... :maybe:

2) Take $ p\equal{}13!k\minus{}13$ for example($ k$ is any positive integer). Than $ p\plus{}1$ is divisible by $ 12$,$ p\plus{}2$ is divisible by $ 11$ and etc. So, none of them is prime. As, there are infinite positive integers, for any positive integer $ k$ we have $ p\equal{}13!k\minus{}13$.

[quote="lasha"]2) Take $ p \equal{} 13!k\minus{}13$ for example($ k$ is any positive integer). Than $ p\plus{}1$ is divisible by $ 12$,$ p\plus{}2$ is divisible by $ 11$ and etc. So, none of them is prime. As, there are infinite positive integers, for any positive integer $ k$ we have $ p \equal{} 13!k\minus{}13$.[/quote]
no,I think he meant that both conditions $ 1,2$ must hold,i.e:
$ p,p\plus{}6,p\plus{}12$ are prime numbers [u]and[/u] $ p\plus{}1,p\plus{}2,p\plus{}3,p\plus{}4,p\plus{}5,p\plus{}7,p\plus{}8,p\plus{}9,p\plus{}10,p\plus{}11$ are composite...

how are you greatestmaths
it easy problem  Albanian eagle (not conjecture, not difficult)
 here is my solution
Let $ n \geq 11$ a positive integer.
let $ u(n)=\frac{n!}{7^{v_{7}(n)}11^{v_{11}(n)}}$ where $ v_{7}(n)$ and $ v_{11}(n)$ are respectively the greatest exposants of 7 and 11 in n.
it obviously that
1) $ u(n)+1$, $ u(n)+7$ and $ u(n)+13$ are primes (for all $ k<u(n)+j^$ where $ j\in{1,7,13}$ we have $ \gcd(k,u(n)+j) =1$)
2)for all $ u(n)+j$ where $ j\in{2,3,4,5,6,8,9,10,11,12}$ we have $ \gcd(j,u(n)+j)=j$
we choose the numbers $ p_n= u(n)+1$.

[quote="NA_219"]how are you greatestmaths
it easy problem  Albanian eagle (not conjecture, not difficult)
 here is my solution
Let $ n \geq 11$ a positive integer.
let $ u(n) = \frac {n!}{7^{v_{7}(n)}11^{v_{11}(n)}}$ where $ v_{7}(n)$ and $ v_{11}(n)$ are respectively the greatest exposants of 7 and 11 in n.
it obviously that
1) $ u(n) + 1$, $ u(n) + 7$ and $ u(n) + 13$ are primes (for all $ k < u(n) + j^$ where $ j\in{1,7,13}$ we have $ \gcd(k,u(n) + j) = 1$)
2)for all $ u(n) + j$ where $ j\in{2,3,4,5,6,8,9,10,11,12}$ we have $ \gcd(j,u(n) + j) = j$
we choose the numbers $ p_n = u(n) + 1$.[/quote]

Unfortunately, you are wrong :

$ n=11$ implies $ u(11)=10!$ and $ u(11) + 7=10!+7$ is not prime (since $ 7$ divides it)

I forgot to say that not for all n hold the first condition holds. In fact it must be that
a-$ u(n)\plus{}1$ is not a multiple of 7
b-$ u(n)\plus{}1$ is not a multiple of 13
c-$ u(n)\plus{}13$ is not a multiple of 7
d-$ u(n)\plus{}7$ is not a multiple of 13
so we must get only the numbers $ n$ which these conditions hold.
If we choose $ n$ such that $ n\equal{}7^{a}13^{b}\plus{}2$ for any strict positive integers $ a$ and $ b$
$ u(n)\equal{}\equiv 2(6!)^{a} \bmod 7$ and $ u(n)\equal{}\equiv 2(12!)^{b} \bmod 13$
so $ u(n)\equal{}\equiv 2(\minus{}1)^{a} \bmod 7$ and $ u(n)\equal{}\equiv 2(\minus{}1)^{b} \bmod 13$.
 We choose  the numbers $ p_n\equal{} u(n)\plus{}1$ where  $ n\equal{}7^{a}13^{b}\plus{}2$ for any strict positive integers $ a$ and $ b$.

excuse me
i make an error ( it's 13 not 11)
Let $ n \geq 13$ a positive integer.
let $ u(n)=\frac{n!}{7^{v_{7}(n)}13^{v_{13}(n)}}$ where $ v_{7}(n)$ and $ v_{13}(n)$ are respectively the greatest exposants of 7 and 13 in n.
it obviously that the following conditions
   1) $ u(n)+1$, $ u(n)+7$ and $ u(n)+13$ are primes (for all $ k<u(n)+j^$ where $ j\in{1,7,13}$ we have $ \gcd(k,u(n)+j) =1$)
    2)for all $ u(n)+j$ where $ j\in{2,3,4,5,6,8,9,10,11,12}$ we have $ \gcd(j,u(n)+j)=j$

hold when
a-$ u(n)+1$ is not a multiple of 7
b-$ u(n)+1$ is not a multiple of 13
c-$ u(n)+13$ is not a multiple of 7
d-$ u(n)+7$ is not a multiple of 13
so we must get only the numbers $ n$ which these conditions hold.
If we choose $ n$ such that $ n=7^{a}13^{b}+2$ for any strict positive integers $ a$ and $ b$
$ u(n)\equiv 2(6!)^{7^{a-1}} \bmod 7$ and $ u(n)\equiv 2(12!)^{13^{b-1}} \bmod 13$
so $ u(n)=\equiv 2(-1)^{ 7^{a-1}} \bmod 7$ and $ u(n)=\equiv 2(-1)^{ 13^{b-1}} \bmod 13$.
 We choose  the numbers $ p_n= u(n)+1$ where  $ n=7^{a}13^{b}+2$ for any strict positive integers $ a$ and $ b$.

i mad an error in the congruence relations

Let $ n \geq 13$ a positive integer.
let $ u(n)=\frac{n!}{7^{v_{7}(n)}13^{v_{13}(n)}}$ where $ v_{7}(n)$ and $ v_{13}(n)$ are respectively the greatest exposants of 7 and 13 in n.
it obviously that the following conditions
   1) $ u(n)+1$, $ u(n)+7$ and $ u(n)+13$ are primes (for all $ k<u(n)+j^$ where $ j\in{1,7,13}$ we have $ \gcd(k,u(n)+j) =1$)
    2)for all $ u(n)+j$ where $ j\in{2,3,4,5,6,8,9,10,11,12}$ we have $ \gcd(j,u(n)+j)=j$

hold when
a-$ u(n)+1$ is not a multiple of 7
b-$ u(n)+1$ is not a multiple of 13
c-$ u(n)+13$ is not a multiple of 7
d-$ u(n)+7$ is not a multiple of 13
so we must get only the numbers $ n$ which these conditions hold.
If we choose $ n$ such that $ n=7^{a}13^{b}+2$ for any strict positive integers $ a$ and $ b$
$ u(n)\equiv 2(6!)^{7^{a-1}13} \bmod 7$ and $ u(n)\equiv 2(12!)^{13^{b-1}7} \bmod 13$
so $ u(n)=\equiv 2(-1)^{ 7^{a-1}13} \bmod 7$ and $ u(n)=\equiv 2(-1)^{ 13^{b-1}7} \bmod 13$.
 We choose  the numbers $ p_n= u(n)+1$ where  $ n=7^{a}13^{b}+2$ for any strict positive integers $ a$ and $ b$.

[quote="NA_219"]i mad an error in the congruence relations

Let $ n \geq 13$ a positive integer.
let $ u(n) = \frac {n!}{7^{v_{7}(n)}13^{v_{13}(n)}}$ where $ v_{7}(n)$ and $ v_{13}(n)$ are respectively the greatest exposants of 7 and 13 in n.
it obviously that the following conditions
   1) $ u(n) + 1$, $ u(n) + 7$ and $ u(n) + 13$ are primes (for all $ k < u(n) + j^$ where $ j\in{1,7,13}$ we have $ \gcd(k,u(n) + j) = 1$)
 ...[/quote]

I think you're wrong.
How could you prove  that $ u(n) + 1$, for example, has no prime divisor $ p$ with $ p > n$ ????

For example, if we take $ a=b=1$ in your proposal, we have $ n=7\times 13+2=93$ and $ u(93)+1=93!+1$ is divisible by $ 271$, and so is not prime.

pco
you are right
maybe i have to search the prime of the form $ n!\plus{}1$
or find another way but i think that this problem has a solution


Prove that $ \forall a_1, a_2, ...a_n \in \mathbb{R}, \exists k \in \mathbb{N^*}$, such that $ \cos{ka_1}, \cos{ka_2}, ..., \cos{ka_n}\geq 0$.

Let $ g(x)\equal{}[4(x\minus{}[x])]$ and $ f(k)\equal{}(g(\frac{ka_1}{2\pi}), g(\frac{ka_2}{2pi}),...,g(\frac{ka_n}{2\pi}))$.
$ f(k)$ had $ 4^n$ means, therefore for $ k\equal{}0,1,...,4^n$ we get $ f(k_1)\equal{}f(k_2)$ and for $ k_0\equal{}|k_2\minus{}k_1|\in N$ we get $ cosk_0a_i\ge 0 \ \forall i$


Find all positive integers $ x,y,z,w$ such that $ w!\equal{}x!\plus{}y!\plus{}z!$.

[quote="moldovan"]Find all positive integers $ x,y,z,w$ such that $ w! \equal{} x! \plus{} y! \plus{} z!$.[/quote]

WLOG, $ x \le y \le z$, then

$ z!<w!\equal{}x!\plus{}y!\plus{}z!<(z\plus{}1)!$ for $ z \ge 3$, impossible.

Case 1: $ z\equal{}1 \Rightarrow x\equal{}y\equal{}1$, which is not a solution.
Case 2: $ z\equal{}2 \Rightarrow 2! <x!\plus{}y!\plus{}2! \le 3!$, equality holds $ \Leftrightarrow x\equal{}y\equal{}z\equal{}2$.

So $ (x,y,z,w)\equal{}(2,2,2,3)$ is the only solution.


I have another 2 questions about determining singularity.

1. What approach should I use to find and classify the singularities (removable / essential / pole) of the function
[e^(z-3) - 1] / [z^3 * (z^2-9)]??

2. Which of the following approach is correct for the function
1 / [1 + 1/(z+1)]??

approach 1: 1 / [1 + 1/(z+1)] = (z+1) / (z+2)
                  So the function has a simple pole at z=-2

approach 2: 1 /[1 + 1/(z+1)] = 1 - 1/(z+1) + 1/(z+1)^2 - 1/(z+1)^3 +...
                  So the function has an essential singularity at z=-1

Thanks for your attention!! :)

For 1), the singularities are $0$, $3$ and $-3$. If you note $f(z)$ your function, then $(z-3)f(z)$ goes to $0$ as $z$ goes to $3$, whence $3$ is removable. The other $2$ singularities will not be removable, and $|f(z)|$ goes to $+\infty$ as $z \rightarrow-3$, whence $-3$ is a pole. Likewise, $0$ is a pole.

For $2$, the singularities are $-2$ and $-1$. $-1$ is removable since $|f(z)|$ has a finite limit at $z=0$ (which is actually $0$), and $-2$ is a pole since $|f(z)|$ goes to $+\infty$ at this point.

Thanks julien_santini for your prompt reply!!  :)  :)

But I have a little puzzle with question 1. Do I need to write the function into a laurent series and check the singularity, or I can ignore the e^(z-3) term and just factorize the denominator??

Well, checking the nature of singularities has (almost) nothing to do with Laurent series ... if $a$ is an isolated singularity of $f$, then only $3$ cases can occur: $lim |f(z)|$ exists and is finite (at point $a$), in which case $a$ is removable; it is $+\infty$, in which case $a$ is a pole, or it does not exist, in which case it is essential. No need to develop anything into Laurent series ... a useful lemma is that $a$ is removable iff $(z-a)f(z)$ tends to $0$ as $z \rightarrow a$.

ic~~ I got it~!~!

Thanks a lot julien_santini!!!  :P


probability
MATHCOUNTS
summer program
MathPath
AwesomeMath
AMC
USA(J)MO

Yeah who do you think will win?

(no voting for yourself :D :P)

Pulak Mittal? Psh.

I think Neal Wu or Nathan Benjamin.

What about Kevin Yang by the way? He got 1st in Cali.

[quote="biffanddoc"]ooo right now I have 100% of the vote :w00tb:[/quote]

What happened to no voting for yourself? I voted for someone else btw.

wait...
WHY THE HECK AM I ON THERE????

Anyways, yeah, you forgot Kevin Yang, and you forgot Kevin Chen too.

I voted for t3h benjamin.

[quote="biffanddoc"][quote="Klebian"]
WHY THE HECK AM I ON THERE????
[/quote]
your not  :huh: :huuh:[/quote]

Pulak Mittal = Klebian.

[quote="biffanddoc"]yeah.
so Klebian (pulak mittal) IS on there.[/quote]

He's arguing with me right now on AIM if he has more than 1% or not, I pointed out guranteed people and likely people and after all those, there's 1 spot left

[quote="biffanddoc"]um how can more than one person be guaranteed to win?[/quote]

Woops, we were talking about countdown

Definitely Nathan. You know what makes him better than Neil. He is an AoPSer.

Nathan left...

Personally, I think there is NO way to predict this.

Throwing up a guess, I would say countdown might look like this (no order):

1. Nathan
2.Neal
3.Biff
4.Kleb
5.Kevin Yang
6.Kevin Chen
7. Me (Allen Yuan)
8. Neil Gurram
9.Eric McCabe
10. Andrew Ardito
11. Edit: Sam Keller (Samath)
12. Edit: Karan Batra

I'll pick Nathan.

[quote="kyyuanmathcount"]Anyone considering me?

I improved 15 points on my chapter score (31 to 46), improved 39 in state (not making to 39) and improved 28 nats (not making to 28), and nat places 229th to 70th.[/quote]
Yes, but if you had taken last years (unofficially) then you would have gotten atleast one right, right? DIdn't you once say "anyone with common sense can get at least a ten..."

Fine, is 29 points not good enough for you?

I also showed improvement

36 to 43 to 44 at Chapters
21st  4th    1st

Not Making it to 40 to 43 at states
-                         1st   2nd

Not Making it to 24 to 35 at Nationals
-                        59th  14th

Countdown is official in our state so I just choked as an 8th Grader at State

I guess those words of encouragement, were really encouraging.

BTW, you only had two individuals from Virginia in the top 57, right?

Yeah, and both made countdown. Jimmy Clark (my roommate) was 58th.

I played both Daniel Li and Jimmy Clark at the countdown at states.  Oddly enough, I beat Daniel, yet was beated by Jimmy.  He was really cool.

[quote="Hamster1800"]Lol i love my improvements:

6th grade: really bad, 30-somethingth at chapter
not too bad, 11th at state

7th grade: 46, 1st at chapter
something decent, 2nd at state
24, 44th at nats

8th grade: 44, 5th at chapter (one target = 4 places :O...way too easy test)
40, 2nd at state (woopie...4th from last year pwned me and 3rd won countdown :()
38, 7th & P at nats (countdown was terrible, being against Ardito....and the same guy pwned me again to get 2nd and Q + masters winner but he lost against Ardito too so I'm ok....especially because my goals were to make countdown and be on TV. I did both and made two newspapers :D)

Secondly, at least two people got 6 individually because Makenzie whateverhislastnameis from the Virgin Islands got a 6 and 227th.

What i find funny is that VA went from 17th to first....How many teams have moved up that much to win in the past? (that was only because the guy who coached our team last year was like "aww it's ok if you get 50th (i think he means team here)...you guys are great for winning state."[/quote]

I'm PRETTY sure somebody got a 5.

It stated clearly on the statistics sheet that the minimum score was a six.

Also, minimum sprint was 2 and minimum target was 0.

However, three people got a 6.

I think we should stop talking about this, in case someone with one of those scores sees this, thanks!

Lol, I doubt the people with insanely low scores would mind too much - I believe the people who did reasonably poorly (58th to 150th place?) that would be the most upset...

Yeah, that converstation with Makenzie Shaw was pretty funny.

"Umm.. guys.. do you know the name of like, that Mexican girl like, from the State Department? She's like, mad hot, can't you get a competitors list on your laptop?"
"Woah man, you type so fast, how do you do it?"
"Listen dudes, if you don't like girls, there's no reason in living, that is your problem right there, not caring about girls"
"Don't you have like, rich parents? Why are you worrying about math?"

I'm guessing he can't be bothered to log on AoPS so he probably won't mind, he got a 4 on the Sprint and 1 on the Target and placed 2nd in the Virgin Islands with a score of EIGHTEEN at states. Since he got 227th, someone had to have a worse tiebreaker than him so I'm guessing 228th had a 2 on Sprint and 2 on Target. If anyone could, would they scan a copy of the distribution sheet? My coach didn't give us a copy and I would like to see it, thanks. ;)

If they got a 6, then they really don't care about mc enough to come to this website...


Determine the smallest positive integer n such that there exist integers a1, a2,……..an
satisfying

(a1)^3+(a2)^3+(a3)^3+...........+(an)^3=2002^2002

[quote]I never finish what i sta :P [/quote]


What can I do if I posted a topic in one part of the forum and i don't get any answers , but i think that if  I would post it in another part of the forum I would get an answer?Am I alowed to double-post?

Yes, that would be fine.  Just try to avoid double posting to begin with.

It is acceptable, provided that some time elapsed (you shouldn't expect immediate answers) but the best thing to do would be to PM a moderator and ask him to move the post. Note that he may disagree with you about whether the post would be more appropriate in that other forum. Anyway, what post are you talking about and where would you like to move it? If it is about mathematics, I'll try to answer :).

Edit: Oops, admins were quicker to respond :)


inequalities
number theory unsolved
number theory

Let $ a_1\equal{}\frac{1}{2}$ and for each n greater than 1,

let  $ a_n\equal{}\sum\frac{1}{d_i\plus{}1}$  where$ d_i$ is the collection of alldistinct postive of n .Prove the inequality   $ n\minus{}lnn<\sum a_i <n$


integration
function
calculus
derivative
calculus computations

The formula for finding an arc length is as follows:

$L = \int_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\ dx$

Find the arclength of the function $y = \frac{x^{3}}{6}+\frac{1}{2x}$, from $x = 1$ to $x = 3$.

First of all, the derivative would be $\frac{x^{2}}{2}-\frac{1}{2x^{2}}$. My question is now, how would you find the arclength? The integral below is correct - but I can't seem to simplify the stuff under the sqrrt..

$L = \int_{1}^{3}\sqrt{1+\left(\frac{(x^{4}-1)^{2}}{4x^{4}}\right)}\ dx$

Keep trying -- it simplifies to a nice perfect square.  (Bringing everything into the fraction should help.)

would it simplify to $\frac{x^{4}+1}{2x^{2}}$?


geometry
circumcircle
trigonometry
geometry unsolved

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14474[/img]

 :(

[img]http://www.mathlinks.ro/Forum/viewtopic.php?mode=attach&id=14491[/img]

 :(

Of course , $ ABC$ is not equilateral  :blush:

Domi

Draw 3 tangents $ a \parallel BC, b \parallel CA, c \parallel AB$ to the given circle $ (I),$ pairwise intersecting at $ A' \equal{} b \cap c, B' \equal{} c \cap a, C' \equal{} a \cap b,$ such that $ (I)$ is incircle of the $ \triangle A'B'C'.$ $ D, E, F$ are feet of perpendiculars from arbitrary $ P \in (I)$ to $ BC, CA, AB$ and $ D', E', F'$ feet of perpendiculars from $ P$ to $ B'C', C'A', A'B'.$ $ \sigma \equal{} PD \plus{} PE \plus{} PF,$ $ \sigma' \equal{} PD' \plus{} PE' \plus{} PF' \plus{} \delta,$ where $ \delta$ is a constant distance independent of $ P.$ $ \sigma, \sigma'$ have min, max for the same positions of $ P.$ Let $ (O')$ be the circumcircle of the $ \triangle A'B'C'.$ According to [url]http://www.mathlinks.ro/viewtopic.php?t=209657[/url], $ \sigma'$ is constant for all triangles with the same incircle $ (I)$ and circumcircle $ (O').$

Let $ A_1, B_1, C_1$ be one of the 2 isosceles triangles with this incircle and circumcircle, such that $ A_1B_1 \equal{} A_1C_1$ and angle $ \angle A_1 \le 60^\circ.$ (The other isosceles triangle has angle $ \angle A_2 \ge 60^\circ.$) Let $ X_1$ be foot of the A-altitude, $ I, O' \in A_1X_1.$ $ A_1X_1 \equiv O'I$ cuts $ (I)$ at $ X_1$ and again at $ X_2.$ $ D_1, E_1, F_1$ feet of normals form $ P$ to $ B_1C_1, C_1A_1, A_1B_1.$ Parallel to $ B_1C_1$ cuts $ A_1X_1$ at $ Q_1.$ $ K_b, K_c$ are feet of perpendiculars from $ X_1$ to $ A_1C_1, A_1B_1.$ $ L_b, L_c$ are feet of perpendiculars from $ Q_1$ to $ A_1C_1, A_1B_1.$ $ M_b, M_c$ are feet of perpendiculars from $ Q_1$ to $ X_1K_b, X_1K_c.$

$ \sigma' \equal{} PD_1 \plus{} PE_1 \plus{} PF_1 \equal{} Q_1X_1 \plus{} Q_1L_b \plus{} Q_1L_c \equal{}$

$ \equal{} Q_1X_1 \plus{} (X_1K_b \minus{} X_1M_b) \plus{} (X_1K_c \minus{} X_1M_c) \equal{}$

$ \equal{} X_1K_b \plus{} X_1K_c \plus{} Q_1X_1 (1 \minus{} 2 \cos B_1)$

$ Q_1X_1$ is the only variable. If $ \angle A_1 < 60^\circ,$ then $ \angle B_1 > 60^\circ,$ $ 1 \minus{} 2 \cos B_1 > 0$ and $ \sigma'$ is strictly increasing as $ Q_1$ moves from $ X_1$ to $ X_2.$ Consequantly, $ \sigma'$ has min/max at the intersections $ X_1, X_2$ of $ (I)$ with the diacentral line $ O'I$ of the $ \triangle A'B'C',$ max at $ X_2$ on the ray $ (IO',$ min at $ X_1$ on the opposite ray.


Determine the number of poker hands of the following tyes:
(a) full houses ( 3 cards of one rank and 2 of a different rank)
(b) straights (5 consecutive ranks)
(c) flushes (5 cards of the same suit)
(d) straight flushes (5 consecutive cards of the same suit)
(e) exactly two pairs ( 2 cards of one rank, 2 cards of another rank, and 1 card of a third rank)
(f) exactly one pair ( 2 cards of one rank, and 3 cards of three other and different ranks)

First off, you can certainly find the solution to this question by searching for "poker" or "poker hand" in the forum.  Second, this question (as well as several others that have been posted in this forum recently by you and chocolatelover) does not belong to "Advanced fields," it belongs to elementary combinatorics.  Future posts of this level would fit much better in the Intermediate forum.

(This thread has been moved by a moderator since this comment.)

Thank you for your comments. I will follow your suggestions.


inequalities
rearrangement inequality
inequalities unsolved

if a,b,c>o, prove that a^3/(b+c) + b^3/(a+c) + c^3/(a+b)>=(a^2+b^2+c^2)/2

although it can be easily solved by rearrangement inequality, i wonder if there is any other method :)

[quote="morheli"]if a,b,c>o, prove that $ a^3/(b\plus{}c) \plus{} b^3/(a\plus{}c) \plus{} c^3/(a\plus{}b)\geq(a^2\plus{}b^2\plus{}c^2)/2$

although it can be easily solved by rearrangement inequality, i wonder if there is any other method :)[/quote]

Here is a solution using Andreescu

$ a^3/(b\plus{}c) \plus{} b^3/(a\plus{}c) \plus{} c^3/(a\plus{}b)\geq(a^2\plus{}b^2\plus{}c^2)/2\Longleftrightarrow$
$ \frac{a^{4}}{a(b\plus{}c)}\plus{}\frac{b^{4}}{b(a\plus{}c)}\plus{} \frac{c^{4}}{c(a\plus{}b)}\geq\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{2}\Longleftrightarrow$

By Andreescu inequality we have

$ \frac{a^{4}}{a(b\plus{}c)}\plus{}\frac{b^{4}}{b(a\plus{}c)}\plus{} \frac{c^{4}}{c(a\plus{}b)}\geq\frac{(a^{2}\plus{}b^{2}\plus{}c^{2})^{2}}{2(ab\plus{}bc\plus{}ca)}\geq\frac{(a^{2}\plus{}b^{2}\plus{}c^{2})(ab\plus{}bc\plus{}ca)}{2(ab\plus{}bc\plus{}ca)}\equal{}\frac{a^{2}\plus{}b^{2}\plus{}c^{2}}{2}$ :)

thank you very much :)
chebysev's inequality seems also working~

BTW,i want to ask how i can type in the way like you, i dun want to use '^', '>=' ....

[quote="galois01"][quote="morheli"]if a,b,c>o, prove that $ a^3/(b \plus{} c) \plus{} b^3/(a \plus{} c) \plus{} c^3/(a \plus{} b)\geq(a^2 \plus{} b^2 \plus{} c^2)/2$

although it can be easily solved by rearrangement inequality, i wonder if there is any other method :)[/quote]

Here is a solution using Andreescu

$ a^3/(b \plus{} c) \plus{} b^3/(a \plus{} c) \plus{} c^3/(a \plus{} b)\geq(a^2 \plus{} b^2 \plus{} c^2)/2\Longleftrightarrow$
$ \frac {a^{4}}{a(b \plus{} c)} \plus{} \frac {b^{4}}{b(a \plus{} c)} \plus{} \frac {c^{4}}{c(a \plus{} b)}\geq\frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{2}\Longleftrightarrow$

By Andreescu inequality we have

$ \frac {a^{4}}{a(b \plus{} c)} \plus{} \frac {b^{4}}{b(a \plus{} c)} \plus{} \frac {c^{4}}{c(a \plus{} b)}\geq\frac {(a^{2} \plus{} b^{2} \plus{} c^{2})^{2}}{2(ab \plus{} bc \plus{} ca)}\geq\frac {(a^{2} \plus{} b^{2} \plus{} c^{2})(ab \plus{} bc \plus{} ca)}{2(ab \plus{} bc \plus{} ca)} \equal{} \frac {a^{2} \plus{} b^{2} \plus{} c^{2}}{2}$ :)[/quote]
Proof 1:
I have :
$ \frac {a^3}{b \plus{} c} \plus{} \frac {a(b \plus{} c)}{4}\geq a^2$
There fore:
$ \frac {a^3}{b \plus{} c} \plus{} \frac {b^3}{c \plus{} a} \plus{} \frac {c^3}{a \plus{} b}\geq a^2 \plus{} b^2 \plus{} c^2 \minus{} \frac {ab \plus{} bc \plus{} ca}{2} \geq \frac {a^2 \plus{} b^2 \plus{} c^2}{2}$
Proof 2 :
By Chebysev inequality and Nesbit inequality we have :
$ \frac {a^3}{b \plus{} c} \plus{} \frac {b^3}{c \plus{} a} \plus{} \frac {c^3}{a \plus{} b} \geq \frac {1}{3}(a^2 \plus{} b^2 \plus{} c^2)(\frac {a}{b \plus{} c} \plus{} \frac {b}{c \plus{} a} \plus{} \frac {c}{a \plus{} b})\geq\frac {a^2 \plus{} b^2 \plus{} c^2}{2}$

[quote="morheli"]i am now trying to break it with sos.

it is equal to show that summation (a-c)( a^2/(b+c) - c^2/(a+b) )>=0
                                   (cyclic)

can anyone help me to prove the upper inequality?[/quote]
 (a-c)(a^2/(b+c)-c^2/(a+b))= (a-c)^2((a^2+c^2+ab+bc+ca)/(b+c)(a+b))>=0 :)

thank you very much :)  now we've got various solution!


inequalities
quadratics
algebra
polynomial
inequalities unsolved

Find the least value of k so that if a,b,c are non-negative real numbers satisfying the condition:[tex]ab+bc+ca=3[/tex] then [tex] (a^2+k)(b^2+k)(c^2+k)\geq\ (1+k)^3[/tex]

This is just brute force, but it's pretty efficient. So let $a+b+c=S$ and $abc=P$ and we will have $S\geq 3$, $P\leq 1$ and $SP\leq 3$. Now, our inequality is

\[(a^2+k)(b^2+k)(c^2+k) \geq (1+k)^3 \Leftrightarrow k^2\sum a^2+k\sum a^2b^2+P^2\geq 3k^2+3k+1 \Leftrightarrow k^2(S^2-9)+k(6-2PS)-(1-P^2)\geq 0\]

I suppose that we are looking for positive values of $k$ (since otherwise $-\infty$ would be a good solution), and note that the above quadratic polynomial in $k$ has one positive root and one negative root. Therefore, that polynomial will be positive iff $k$ is greater than the positive root, i.e.

\[k\geq \dsp \frac {PS-3+\sqrt {P^2S^2-6PS+9+S^2-9-P^2S^2+9P^2}}{S^2-9}=\frac {PS-3+S-3P}{S^2-9}=\frac {1+P}{S+3}=\frac {P+P^2}{SP+3P}\geq \frac {P+P^2}{3+3P}=\frac P3\]

and since we can make $P$ as close to 1 as we want to, I guess that the minimum value of $k$ which always works is $\dsp \frac 13$.


Tina left her office at some time after 2pm. When she returned later the minute and hour hands on her office clock had swapped positions.

At what time, to the nearest minute, did Tina leave her office?

Source: NSW Math Competition 2005

are you sure you gave enough info.............

I copied it word for word from the source, so the answer is yes. And it's very possible to solve it cause I solved it after making a mistake the first time.  :)

Sorry aidan, you are right, I just didn't get it at the first look at it

My first guess 2:16.......?

At around 2 o'clock , say the minute hand move $k$ minute after $2$pm , the hour hand move $k/12+10$ from 12 noon .

At around 3 o'clock , the minute hand move $j$ minute after $3$ pm , the hour hand move $j/12+15$ from 12 noon .

since the hour hand and minute hand swap place , we have simultaneous equation

$k=j/12+15$
$j=k/12+10$

solve and get $k=15\frac{135}{143}$ which is about $16$ minute pass $2$ .

[quote="aidan"]Excellent solutiong shyong. Anyway, I have a gut feeling you did this question last year eh?  :lol:[/quote]

when the first time i do this question , i stuck and just simply tick an answer and that was my lucky day  :)

After i went back and study the problem , and I saw that we can do it in such way .Another type of this " minute hour hand" problem is time when hour and minute hand overlap one another , time when the hour hand and minute hand make an angle $\theta$ . So basically they can be solved by similar method .

Wow, not so fair eh? I guessed wrongly and you guessed correctly. And you're right about the other minute-hour hand problems... how about this extension question (which is easier than the previous one)?

How many minutes after 6 pm do the hour hand and minute hand coincide between 6pm and 7pm?

in general , for this type of problem , the two simultaneous equation is

$k=\frac{j}{12}+5a$
$j=\frac{k}{12}+5b$

So $k=\frac{60(b+12a)}{143}$ , $j=\frac{60(a+12b)}{143}$ .

 :)

As for the overlapping of hour and minute hand , just solve the equation

$k=\frac{k}{12}+5a$ $\implies k=\frac{60a}{11}$

where $k$ is minute pass 12 and  $a$ is the hour you want . So to answer you question , pick $a=6$ ;)

Lol.. a bit too easy for you eh. That generalisation is interesting.

Yeah that generalization is pretty interesting.
My guess involved pure logic instead of mathematical concept, but these are quite nice problems


Let $ a_{1}, a_{2}, \ldots, a_{n}>0$. Prove that the following inequality holds:

  \[ \prod_{j\equal{}1}^{n}\left(\sum_{i\equal{}1}^{n}a_{ij}^{m}\right)\geq\left(\sum_{i\equal{}1}^{n}\prod_{j\equal{}1}^{m}a_{ij}\right)^{m}.\]

I think it is :$ \prod^m_{j\equal{}1}\sum^n_{i\equal{}1} a^m_{ij}\geq(\sum^n_{i\equal{}1}\prod^m_{j\equal{}1} a_{ij})^m$

$ A_j\equal{}\sum_{i\equal{}1}^n a_{ij}^m$ and $ x_{ij}\equal{}A_j^{\minus{}\frac{1}{m}} a_i$
We see that
$ \sum _{i\equal{}1}^n x_{ij}^m\equal{}1$ for every j. Hence using AM-GM
$ \sum_{i\equal{}1}^n x_{i1}x_{i2}...x_{ij}\leq 1\iff (\sum_{i\equal{}1}^n a_{i1}a_{i2}...a_{ij})^m\leq A_{1}A_{2}....A_{j}$ QED


Solve this classic.
$60$ is the Circumference, $2$ is the perpendicular, find the chord.

Basically there is a circle of $60$ usints in circumference. Draw a chord in the circle. The perpendicular to the chord from a point on circle is $2$ units. What is the length of the chord?

What's "Classic" about this problem?

Just draw the triangle, if the perpendicular bisects the segment.

If not, then the problem statement is unclear; where is this perpendicular? (We could say the perpendicular from the circle to the chord approaches zero as we approach one of the endpoints.)

[quote="cincodemayo5590"]What's "Classic" about this problem?

Just draw the triangle.[/quote]

"classic" is that its very old problem, abt 2000 years old or so... may be older. thats all.

Um... ok. Sure.

We still need to know where the perpendicular hits the chord.

[hide="If it bisects, just use"]
pythagorean.[/hide]

[quote="xxxyyyy"]$60$ is the Circumference, $2$ is the perpendicular, find the chord.[/quote]
[hide="Am I right?"]The radius is $\frac{30}{\pi}$. Hence, we have a right triangle whose hypotenuse is the radius and whose leg is $\frac{30}{\pi}-2$. The other leg is half the length of the chord. So the answer is:
$(1/2) \sqrt{\left( \frac{30}{\pi}\right)^{2}-\left( \frac{30}{\pi}-2 \right)^{2}}$[/hide]
To a mod: Move to HSB.


Given a right-angled triangle with integral side lengths $ a,\ b,\ c\ (a<b<c)$.
Prove that if $ a$ is odd, then $ b$ is a multiple of 4.

[quote="kunny"]Given a right-angled triangle with integral side lengths $ a,\ b,\ c\ (a < b < c)$.
Prove that if $ a$ is odd, then $ b$ is a multiple of 4.[/quote]

If $ b$ is odd, then $ c^2\equal{}a^2\plus{}b^2\equal{}1\plus{}1\equal{}2(mod 4)$, impossible, so $ b$ is even and $ c$ is odd.

Now $ b^2\equal{}c^2\minus{}a^2\equal{}1\minus{}1\equal{}0(mod 8)$, which leads to $ 4|b$

It is well known that the sides of a right triangle can be notated as following:
The legs are $ r^2\minus{}s^2$ and $ 2rs$ and the hypotenuse is $ r^2\plus{}s^2$, where s is always even.  Since once leg is odd, it must be the $ r^2\minus{}s^2$ leg.   Since $ s|2$ we know $ 2rs|4$, we are done.

[quote="kunny"]Given a right-angled triangle with integral side lengths $ a,\ b,\ c\ (a < b < c)$.
Prove that if $ a$ is odd, then $ b$ is a multiple of 4.[/quote]

this is obvious: the sides of an iteger sided right triangle are of the form $ 2m^2n^2,m^2\minus{}n^2,m^4\plus{}n^4$.since $ m^2\minus{}n^2$ is odd, one of $ m,n$ is even and the result follows.

[quote="Arrange your tan"][quote="Arrange your tan EDIT"][quote="kunny"]Given a right-angled triangle with integral side lengths $ a,\ b,\ c\ (a < b < c)$.
Prove that if $ a$ is odd, then $ b$ is a multiple of 4.[/quote]

this would follow: the sides of an integer-sided right triangle are of the form $ 2mn,m^2 \minus{} n^2,m^2 \plus{} n^2$, since $ m^2 \minus{} n^2$ is odd, one of $ m,n$ is even and the result follows.[/quote]

Your exponents were too high for the first and the third of your listed expressions
for the sides, for one thing.[/quote]

sorry for my little mistake.


AMC USA(J)MO USAMO geometry incenter circumcircle LaTeX

ABC is a triangle with incenter I. Show that the circumcenters of IAB, IBC, ICA lie on a circle whose center is the circumcenter of ABC.

 ABC is acute-angled. The circle diameter AB meets the altitude from C at P and Q. The circle diameter AC meets the altitude from B at R and S. Show that P, Q, R and S lie on a circle(centered at A in fact).

hints:

[hide="first"]Show that the three circumcenters are equidistant from O by proving isosceles triangles (by angle chasing)

Edit--paladin8 essentially did what I had in mind.[/hide]
[hide="second"]Power of a Point![/hide]

[hide="first"]
Let the circumcenters of $ICA$, $IAB$, and $IBC$ be $W$, $Y$, and $Z$, respectively, and the circumcenter of $ABC$ be $O$ (Note: X in LaTeX shows up as a flag on my computer when by itself... $X$)

We know the circumcenter is the intersection of the perpendicular bisectors, so call $D$ the midpoint of $AZ$ and $E$ the midpoint of $AC$ ($E$ is on $OW$ because $O$ and $W$ are both on the perpendicular bisector of $AC$). Let $F$ be the intersection of $AC$ and $DW$. Since $\angle ADF = \angle WEF = 90$ (by perpendicular bisectors) and $\angle AFD = \angle FWE$, we have $ADF ~ WEF$ so $\angle OWF = \angle EWF = \angle DAF = \angle IAF$.

Similarly, denote $G$ the midpoint of $AB$ and $H$ the intersection of $AB$ and $DY$. Then we have $\angle OYH = \angle GYH = \angle DAH = \angle DAF = \angle IAF$ because the incenter is the intersection of the angle bisectors. So $OWY$ is isosceles and $OW = OY$.

Do the same for $OYZ$ or $OZW$ and combine them to get $OY = OW = OZ$, so $W$, $Y$, and $Z$ are on a circle with center $O$.[/hide]


linear algebra
matrix
combinatorics unsolved
combinatorics

Let $ A_1,A_2,...,A_m$ be distinct subsets of a set $ A$ with $ n\geq 2$ elements. Suppose any two of the subsets have exactly one element in common. Prove that $ m\leq n$.

Classical linear algebra problem applied to combinatorics (and probably posted before).

Let $ x_1, \cdots, x_n$ be the elements.
Consider the $ n*m$ matrix $ M\equal{}(a_{i,j})$ where $ a_{i,j} \equal{} 1$ if $ x_i \in A_j$ and $ a_{i,j} \equal{} 0$ otherwise.
Let $ M^t$ be the transposed matrix of $ M$.
Then $ P\equal{}M^t*M$ is an $ m*m$ matrix and its coefficient $ m_{i,j}$ equals the cardinality of the intersection of $ A_i$ and $ A_j$.
Thus, $ m_{i,j} \equal{} 1$ if $ i \neq j$ and at least one of the $ m_{i,i}$ is not equal to $ 1$ otherwise two of the sets would be equal. It follows that $ P$ is the matriw of a nondegenerated bilinear positive form, thus $ P$ is inversible and has rank $ m$. Thus, $ M$ has rank at least $ m$, so that $ m \leq n$.

Pierre.


trigonometry calculus derivative function parameterization logarithms algebra

Find the general solution of: $ y'' \plus{} y' \plus{} a^2e^{2x}y\equal{}0,$ where $ 0 \neq a \in \mathbb{R}$ is a constant.

[hide="answer"]
Let $ t\equal{}ae^x$ and $ v(t)\equal{}ty(x)$, then it's a routine calculation to show that $ v(t)$ satisfies $ v''(t)\equal{}\minus{}v(t)$ so $ v(t)\equal{}c_1\sin t \plus{}c_2 \cos t$, and therefore \[ y(x)\equal{}e^{\minus{}x}\left(c_1\sin ae^x \plus{} c_2\cos ae^x\right)\]
[/hide]

[quote="JoeBlow"][hide="answer"]
Let $ t \equal{} ae^x$ and $ v(t) \equal{} ty(x)$, then it's a routine calculation to show that $ v(t)$ satisfies $ v''(t) \equal{} \minus{} v(t)$ so $ v(t) \equal{} c_1\sin t \plus{} c_2 \cos t$, and therefore
\[ y(x) \equal{} e^{ \minus{} x}\left(c_1\sin ae^x \plus{} c_2\cos ae^x\right)
\]
[/hide][/quote]

I think by $ v''(t)$ you mean the second (partial) derivative of $ v(t) \equal{} ty(x)$ with respect to $ t$.  If so, $ v''(t)$ should be 0 if I'm not wrong, but $ \minus{}v(t) \equal{} \minus{}ty(x) \neq 0$.  Am I missing something?  :maybe:

No, there is no partial differentiation here, $ v$ and $ y$ are both functions of one parameter, but I've defined a new parameter $ t$ as a function of $ x$.  So really $ v(t) \equal{} ty(x) \equal{} ty\left(\log \frac {t}{a}\right)$ and hence
\[ v''(t) \equal{} \frac {1}{t}\left(y'\left(\log \frac {t}{a}\right) \plus{} y''\left(\log \frac {t}{a}\right)\right)
\]
But from the ODE we see that $ y' \plus{} y'' \equal{} \minus{} a^2e^{2x}y \equal{} \minus{} t^2y \equal{} \minus{} tv(t)$, so $ v''(t) \equal{} \minus{} v(t)$.  Of course you need to be careful about properly defining the domain of $ t$ depending on the sign of $ a$, but the answer comes out the same for each case.


trigonometry
function
calculus
calculus computations

Solve the following equations:

[size=150]1.[/size]

$(x\cos y-y\sin y)dy+(x\sin y+y\cos y)dx=0$

[size=150]2.[/size]

$\left(x^{4}y^{2}-y\right)dx+\left(x^{2}y^{4}-x\right)dy=0$

[size=150]3.[/size]

$e^{x}(x+1)dx+\left(ye^{y}-xe^{x}\right)dy=0$

well these are all differential equations in the form:

$M(x,y)dx+N(x,y)dy=0$

where they may be exact equations, and if they're not then we can make them an exact equation by multiplying a function $\mu$ to the equation

where $\mu$ will be found by the following:

$\frac{1}{\mu}=\frac{N_{x}-M_{y}}{M\mu_{y}-N\mu_{x}}$

where by $M_{y}$ we mean $\frac{\partial M}{\partial y}$.

for the first equation:

$M_{y}=x\cos y+\cos y-y\sin y$
$N_{x}=\cos y$

so $N_{x}-M_{y}=y\sin y-x\cos y$

and so by choosing $\mu(t)=\mu(x)$ then:

$\mu_{y}=\frac{d\mu}{dt}\frac{\partial t}{\partial y}=0$
$\mu_{x}=\frac{d\mu}{dt}\frac{\partial t}{\partial x}=\frac{d\mu}{dt}$

and:  $\frac{d\mu}{\mu}=\frac{N_{x}-M_{y}}{-N}=1dt \rightarrow \mu=e^{x}$

so by multiplying $e^{x}$ to the equation it'll turn into an exact differential equation and...


Let $\mu_{i}$, $i=1,2,...,n$ be mutually non-singular non-atomic measure defined on the space $X$ and $\forall i$, $\mu_{i}(X) = 1$.Prove that for any integer $k \geq 1$ there exists a subdivision $X = \bigcup_{i=1}^{k}X_{i}$ such that $\mu_{i}(X_{j}) = \frac{1}{k}$ for all $1 \leq i \leq n$ and $1 \leq j \leq k$.

What if all the $\mu_{i}$ are all equal to the Dirac measure at some point of $X$? The conclusion fails to hold for trivial reasons: the measures $\mu_{i}$ only take the values $0,1$.

Am I missing something obvious, or did you forget to mention some conditions?

I added the non-atomic condition to the problem, hope it will work in this case.
Sorry for unconvenience.

Sorry.Here i hope is the right formulation of the problem:

Let $\mu_{i}, i=1,2,...,n$ be pairwise not-absolutely continuous measures defined on the space $X$ and $\forall i$, $\mu_{i}(X) = 1$.Prove that for any integer $k \geq 1$ there exists a subdivision $X = \bigcup_{i=1}^{k}X_{i}$ such that $\mu_{i}(X_{j}) = \frac{1}{k}$ for all $1 \leq i \leq n$ and $1 \leq j \leq k$.

the counter-example of Grobber still works  :maybe:

Oh,yes, i missed the word "not", not-absolutely continuous.


In how many ways can a board $ 6 \times 6$ be filled with pieces $ 3 \times 1$ or $ 1 \times 3$?

[hide="remark"]
I've already counted $ 72$ but maybe there are more.[/hide]

I suspect that $ A_n$, the number of ways to tile a $ 6 \times n$ board, has a rational generating function.  Unfortunately, it doesn't seem to be very easy to calculate.  Are you familiar with the corresponding problem for $ 1 \times 2$ and $ 2 \times 1$ pieces?

Well, it's certainly rational because we can write down a system of linear recursions with it as one of the solutions.  (Just write one for every possible shape that the left edge might take after we've put down some bricks.)  This is, however, extremely unpleasant to do in practice.


Dear Mathlinkers,
let ABC be a triangle, (1) the incircle, DEF the contact triangle of ABC, Hc the orthocenter of DEF, U the antipole of D wrt (1), and U', U" the second points of intersection of AU, UHc with (1).
Prouve: U'U" is parallel to EF
Sincerely
Jean-Louis

Dear Jayme,
You can see my post at here:
http://www.mathlinks.ro/viewtopic.php?p=1560316&sid=9af1b0f651a8323f33d72f89cd650667#1560316
It's also a well-known result about harmonic quadrilateral.

Dear Livetolove and Mathlinkers,
my proof is based on UA which is the U-symmedian of the triangle UEF
and on UHc that is the U-median of UEF (that you have mentioned)... after we are done.
Sincerely
Jean-Louis


function
search
real analysis
real analysis unsolved

Does there exist a sequence of positive continuous functions on $ \mathbb{R}$ such that $ \{f_n(x)\}$ is unbound if and only if $ x$ is irrational.

The search button is your friend ;). Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=299507]here[/url] for instance

Thank you very much,And another two problems:

1.Does there exist a sequence of positive continuous functions on  $ \mathbb{R}$ such that $ f_{n}(x)\to \infty$ if and only if  $ x$ is irrational?

Does the sequence of positive continuous functions your write such that $ f_{n}(x)\to \infty$ for every $ x$ is irrational?
\[ f_{n}(x) \equal{} \min_{1\le k\le n}(k \plus{} n|x \minus{} q_{k}|)\,,\]
2.Does there exist a sequence of positive continuous functions on  $ \mathbb{R}$ such that $ f_{n}(x)\to \infty$ if and only if  $ x$ is rational?

[quote="zhaobin"]Thank you very much,And another two problems:

1.Does there exist a sequence of positive continuous functions on  $ \mathbb{R}$ such that $ f_{n}(x)\to \infty$ if and only if  $ x$ is irrational?

Does the sequence of positive continuous functions your write such that $ f_{n}(x)\to \infty$ for every $ x$ is irrational?
\[ f_{n}(x) \equal{} \min_{1\le k\le n}(k \plus{} n|x \minus{} q_{k}|)\,,\]
[/quote]
Oops.I think in your example we can prove that  $ f_{n}(x)\to \infty$ if and only if  $ x$ is irrational.

So can somebody tell me the answer of the last questions of this type.

2.Does there exist a sequence of positive continuous functions on  $ \mathbb{R}$ such that $ f_{n}(x)\to \infty$ if and only if  $ x$ is rational?

The search button is your friend ;). Look [url=http://www.artofproblemsolving.com/Forum/viewtopic.php?t=213386]here[/url] for instance.

Thank you very much :) And I have collect some problems like this in my blog. :P


No matter which 55 positive integers may be selected from (1, 2,… 100), prove that you must choose some two that differ by 9, some two that differ by 10, some two that differ by 12, and some two that differ by 13, but that you need not have any two that differ by 11.

We restate the problem as follows.
$ D$ is a subset of $ \{1,2,...,100\}$, with $ |D|\equal{}55$. Prove that for $ n\equal{}9,10,12,13$, there exist two elements of $ D$ that differ by exactly $ n$, and that this is not necessarily true if $ n\equal{}11$.

For $ n\equal{}11$ we have this counterexample: $ D\equal{}\{d, d\plus{}22, d\plus{}44, d\plus{}66, d\plus{}88  |  d\equal{}1,2,...,11\}$.

For $ n\equal{}9,10,12,13$ we will use very similar arguments, so we'll only show the first case.
We divide $ \{1,2,...,100\}$ in $ n$ subsets acoording to the remainder of the numbers when dividing by $ n$. For $ n\equal{}9$ these are:
$ 1,10,19,...,100; 2,11,20,...,92; 3,12,21,...,93; ...; 9,18,27,...,99$.

If we want to construct some $ D$ for which what we are trying to prove is false, $ D$ can't contain two "consecutive" numbers of one of these subsets. So, if a subset has $ m$ elements, $ D$ contains not more than $ \left\lceil \frac{m}{2} \right\rceil$ elements of that subset.

The first subset has $ 12$ elements and all the others have $ 11$ elements. So $ D$ can't contain more than $ 6$ elements from each subset $ \rightarrow$ $ D$ can't contain more than $ 6 \times 9 \equal{} 54$ elements in total.


algebra
polynomial
abstract algebra
number theory unsolved
number theory

Prove that for any positive integer $n$, the number $1+5^{n}+5^{2n}+5^{3n}+5^{4n}$ is not a prime.

[quote="sxcxc"][quote="Rust"]Let $m=ord_{5}(n)$, then $n=k5^{m}$.
It give
$\frac{5^{m+1}-1}{5^{m}-1}|1+5^{n}+5^{2n}+5^{3n}+5^{4n}=\frac{5^{5n}-1}{5^{n}-1}.$
[/quote] Can you explain that in more detail?[/quote] He meant that $5^{m}|n$ but $5^{m+1}\not | n$ (this isn't order, you don't write this as $ord_{5}(n)$, but, if I am not wrong, as $v_{5}(n)$).

I just want to show a way for generalization of this problem. Well it's not really a generalization, but, oh just look down.

If we put that $P_{k,n}(x)=1+x^{n}+x^{2n}+...+x^{(k-1)n}$, in our problem we want to examine whether $P_{5,n}(5)$ is prime.

[color=blue]Lemma: If $k$ is prime, and $P_{k,n}(x)$ is prime for some $x$, then $n=k^\alpha$.[/color]

Proof: We will show that $P_{n,k}(x)$ iz reducible over $Z[x]$ when $n$ isn't a power of $k$.
 Look at polynomial $P_{1,k}(x)$. All its roots are $k$th roots of unity, except $1$. If $\epsilon$ is one its root, then all the others are given with $\epsilon, \epsilon^{2},...,\epsilon^{k-1}$ and $\epsilon^{k}=1$.
Now, if $k\not |n$ then $\{1,2,...,k-1\}=\{n,2n,...,(k-1)n\}$ in module $k$, so $0=1+\epsilon+\epsilon^{2}+...+\epsilon^{k-1}=1+\epsilon^{n}+\epsilon^{2n}+...+\epsilon^{(k-1)n}$, which implies that every root of $P_{1,k}(x)$ is also a root of $P_{n,k}(x)$, so $P_{1,k}(x)|P_{n,k}(x)$.
When $k|n$ let $n=k^\beta \cdot m$ where $k\not |m$ and $m>1$.
Now, because of that what we proved before, it stands that $P_{1,k}(x^{k^\beta})|P_{m,k}(x^{k^\beta})=P_{n,k}(x)$, and $m>1$ so $deg(P_{1,k})<deg(P_{m,k})$.

The conclusion: When $n$ isn't a power of $k$ then polynomial $P_{n,k}(x)$ is reducible (over $Z[x]$). So, if we want $P_{n,k}(x)$ to be prime for some $x$ then $n$ must be power of $k$.

Bye


In an $ n\times n$ array, each of the numbers $ 1,2,...,n$ appear exactly $ n$ times. Show that there is a row or a column in the array with at least $ \sqrt{n}$ distinct numbers.


To Valentin:

Can you increase my upload quota from 10 MB to something higher? I need to upload an important document for AMC forum, and it says that I reached my limit. Thanks.

Also, I want to point out possible "bug." When I click on Attachment in my Control Panel, it says that no options are available. But isn't the purpose of Attachment to see what attachments you made and delete them (without having to go through individual thread)? Please fix this.

It's same regardless of skin I am in (MathLinks or AoPS). Thanks.

I don't know why but attachments always seem to be problem for me...  :|


Find [tex]\displaystyle \sum_{x=0}^{n} \left(\begin{array}{cc}n\\x\end{array}\right)^2[/tex]

The book had a really smooth solution to this.

This is really well-known: We write it as $\displaystyle \sum_{i=}^n\binom ni\binom n{n-i}$. This is the coefficient of $x^n$ in $(1+x)^{2n}$ and, at the same time, that coefficient is $\displaystyle \binom {2n}n$, so that's the answer.

grobber, this time I think I have a much less elegant solution:

We play around with small values of n and get a feel that the answer should be (2n C n). But once we have that "guess", the proof turns out to be ridiculously easy--consider the number of ways to select n objects from 2n objects. One way is (2n C n), the other is to divide the 2n objects first into 2 subsets (each containing n), then we just have to choose k from one subset and (n-k) from the other. Sum up all k to get the identity. :D

okay, I see I have changed the variables... :?  so it's:

Sum (n C k)=(2n C n)... :lol:


group theory
abstract algebra
superior algebra
superior algebra unsolved

1. Suppose $ |G|\equal{}p^m$, where $ p$ is a prime and $ m$ is a positive integer. Prove that there exists $ H<G$ with $ |H|\equal{}p$.

2. Suppose $ H$ and $ K$ are two finite subgroups of $ |G|$. Let $ HK\equal{}\{hk: h\in H, k\in K\}$, show that $ |HK|\equal{}\frac{|H|\cdot|K|}{|H\cap K|}$.

Can you use Cauchy's Theorem for 1? If so, pick the element of order $ p$ and look at the subgroup that it generates.

For 2: This has been asked twice in the past two or three weeks, so it shouldn't be too far down. I believe it was proved in a post titled "no simple group 180" or something like that.

[quote="HilbertThm90"]Can you use Cauchy's Theorem for 1? If so, pick the element of order $ p$ and look at the subgroup that it generates.[/quote]
It [i]is[/i] Cauchy's Theorem, so I hardly see how you can use it. :)

The proof I like best is to look at the set of $ p$-tuples of elements of $ G$ whose cyclic product is $ 1$, and consider the action of the cyclic group of order $ p$ by cyclic permutation.  You can show that this set has order divisible by $ p$, so the number of fixed points is divisible by $ p$.  But the fixed points correspond exactly to elements of $ G$ of order dividing $ p$, and $ 1$ is already one of those.

You both exagerate it a bit too much  :wink:
Just take any nontrivial element $ x$, it will have order $ p^k$. Now look at $ x^{p^{k \minus{} 1}}$.

Edit: see below.

We seem to have both overlooked the condition on $ G$.  :blush:

Hmmm, i think ZetaX proof is incorrect.

Consider the $ G\equal{}\mathbb{Z}_2 \times \mathbb{Z}_4$. So $ |G|\equal{}2^3$.

Now, say we pick the non-trivial element $ (0,1)$. It is not of order $ 2^3\equal{}8$, instead the order is $ 4$.

Yes, $ 8(0,1)\equal{}(0,0)$, but $ 4(0,1)\equal{}(0,0)$ too and $ (0,0)$ does not generate a subgroup of order $ 2$.

I think he meant for the $ m$ to be different than the $ m$ in the problem.  Yes, the order may be $ p^k$ with $ k<m$, but this doesn't break the proof.

Sorry, thought the problem said $ |G|\equal{}p^n$ (so $ m$ is indeed meant to be something different).


If two faces of a dice have a common edge, the two faces are called adjacent faces. In how many ways can we construct a dice with six faces such that any two consecutive numbers lie on two adjacent faces?

$\textbf{(A)}\ 10  \qquad\textbf{(B)}\ 14  \qquad\textbf{(C)}\ 18  \qquad\textbf{(D)}\ 56  \qquad\textbf{(E)}\ \text{None}$

[hide="Solution"]Let's place $ 1$ and $ 2$ to the dice. Now open the dice like the shape below:
[code]
  1
  2
A B C
  D
[/code]

If $ A\equal{}3, B\equal{}4$, $ C$ and $ D$ can be either $ 5$ or $ 6$. There are $ 2$ ways.
If $ A\equal{}3, D\equal{}4$, $ B$ and $ C$ can be either $ 5$ or $ 6$. There are $ 2$ ways. So if $ A\equal{}3$, there are $ 4$ ways.
Similary if $ C\equal{}3$, there are $ 4$ ways.
If $ B\equal{}3$, $ D$ cannot be $ 4$ because $ A$ and $ C$ are not adjacent. So if $ A \equal{} 4$, $ D$ must be $ 5$ and $ C$ must be $ 6$. if $ C\equal{}4$, then $ D\equal{}5, A\equal{}6$. So there are $ 2$ ways.
Totally there are $ 4\plus{}4\plus{}2\equal{}10$ ways to construct this dice. $ \boxed A$[/hide]


Anyone else going? I got 14th at Cincinnati, so I barely got in...

My little bro is going.  He got 4th at chapter, not sure exactly the area it covers, but we live in Mount Vernon...

13th in Cincinnati. :D

It's my 1st year doing MathCounts.. so I don't think it's too shabby.

Pii, do you remember me? I was the other Indian kid who was sitting in the front row besides Vadadi from Sycamore. I lost in the 1st round of the countdown :P

Who were you?

I never really knew about math counts until this year, and I'm a junior, so [hide="..."] :whistling:  :noo:  :dry: [/hide]

Yeah, I didn't learn about it until last year.  This is the first year we've had it in Mount Vernon, and I've been the coach...

I lost in the 1st round of countdown too... I was sitting behind the timer. It's my second year and my goal was to get into the countdown, so I made it. Yay. :lol:

What!?!  You two are from Cincinnati??

I was in that chapter last year when I did Mathcounts.

I lost in the 2nd round in every countdown I participated in (1 away from trophy) except for 2006 state.

Hmm... I was really nervous and every second I was taking to check my work, felt like giving an hour for my opponent to buzz in... so I just buzzed in and got the answer barely off both times.. the second one was even a calculation mistake :P

Basically, my opponent didn't really win, I lost for him :lol:

Anyway... it doesn't matter too much since the Countdown is unofficial again in the state. :)

I really want to get in the Nationals, but I have no chance- top 4 is just too much, Sycamore will dominate probably... take up all 4  :(

You might surprise yourself.  People have gotten a higher placing at state than chapter.

I stopped the Sycamore sweep last year.  They took 1,2,4,5 and I took 3.  I think Sycamore is the better school so I am not proud of it.

Hmm.. well I hope thats my case. I REALLY want to get in to the state.. I don't really care what happens to the nationals :P

Let's see how it goes....

*goes back to doing sprints*

ehehheehee: :clap:  :clap:  :clap: for stopping the Sycamore sweep!
Yeah, I went too fast and...yeah. I was the one that accidentally buzzed in twice... :blush: oops.
I don't care what happens at nationals, either, as long as I get there.

ehehheehee makes a very valid point - case in point: me. I went to nationals twice (2005 and 2006), but I never placed in the top 5 at the Cincy chapter.

I got the cutoff for nationals on the sprint for last year- just did it last night.

I'm feeling a bit better now :P

So... it's this Saturday......any comments from anyone??

Did Sycamore take all of the top spots again?

Eh, I'm not even from Ohio but look around first...  Sycamore
Jr. High only has one person and the coach going...  Link:
http://www.mathcounts.org/webarticles/articlefiles/815-State_Winners_07.pdf

Should have known that... thanks though.


Prove that there exist a natural number $ n$,such that among the numbers
$ n\plus{}1,n\plus{}2,\dots,n\plus{}2008$
there is exaclty $ 8$ prime numbers

As you increase $ n$ by one the number of primes in this interval changes exactly by $ 0$ or $ \pm1$

If we define $ f(n)\equal{} (\text{number of primes in n\plus{}1, n\plus{}2..... n\plus{}2008})$
then $ f(n)$ changes exactly by $ 0$ or $ \pm1$ as we go up. Now $ f(0)> 8$ certainly, and $ f(2009!\plus{}1)\equal{}0$

and so somewhere anlong the way we must have $ f(n) \equal{} 8$


analytic geometry
combinatorics proposed
combinatorics
cono sur

We divide the plane in squares shaped of side 1, tracing straight lines parallel bars to the coordinate axles. Each square is painted of black white or. To each as, we recolor all simultaneously squares, in accordance with the following rule: each square $Q$ adopts the color that more appears in the
configuration of five squares indicated in the figure. The recoloration process is repeated indefinitely.

Determine if exists an initial coloration with black a finite amount of squares such that always has at least one black square, not mattering how many seconds if had passed since the beginning of the process.

The problem is incomplete.

Part b) ¿Is there any configuration of $n$ black squares such that after some finite time there are at least $10^{10}n$ black squares?

The answer to part (a) is that there is no configuration that will continue indefinitely.  Consider the convex hull of the centers of the black squares.  No black square can be generated beyond the convex hull because that would have implied the angle there would have been concave (contradicting that it is convex).  Then, any black square on a "corner" of the convex hull will be eliminated because it must have only $4$ or less black squares next to it.  So, the convex hull will get smaller every iteration of recoloring and eventually no black squares will exist.


Gregor is planting pea plants in his backyard.  Some of the plants are tall and the rest are short. Some of the plants are yellow and the rest are green.  The numbers of different plant types are in the ratio \[ \text{Tall green}: \text{Tall yellow}: \text{Short green}: \text{Short yellow} \equal{} 8: 5: 2: 1.\] If there are 1080 green plants, then how many total plants are in Gregor's yard?

Let's set the number of short, yellow plants as x.
Then, there are $ 8x \plus{} 2x \equal{} 10x$ green plants.
$ 10x \equal{} 1080$
$ x \equal{} 108$
There are $ 8x \plus{} 5x \plus{} 2x \plus{} x \equal{} 16x$ total plants.
Therefore, There are $ 16 * 108 \equal{} \boxed{1728}$ total plants.


3 cards are chosen at random from a standard 52-card deck.  What is the probability that they form a pair?  (A 3-card hand is a `pair' if two of the cards match in rank but the third card is different.  For example, 668 is a pair, but 999 is not.)

We choose a rank, two of the four suits of that rank, and then another rank, which we have four choices of, so the probability is

$ \frac{13\dbinom42\cdot12\dbinom41}{\dbinom{52}3}\equal{}\boxed{\frac{72}{425}}$.

I can understand the solution of math154, however I couldn't figure out what was wrong in the following argument:

I first randomly choose a card from the 52 card deck. There are 52 ways. Then I choose a pair for that card that I just picked. There are only 3 ways to do that. For example, if the first card I picked is Ace of Hearts, then I can choose one of other Aces (three Aces remaining). The 3rd card has 48 ways to choose using the exact same way as math154.

However, this results in a different answer:
\[
\frac{52\cdot 3 \cdot 12\dbinom{4}{1}}{\dbinom{52}{3}} = \frac{144}{425}=2 \cdot \frac{72}{425}.
\]

 What is wrong with the above argument? I have an "answer" that is twice of the correct answer.

It seems that I over-counted. But I do not see what I did wrong.

Now, I realize what I did wrong. The pair chosen by my method is double counted. The Ace of Hears chosen first then forms a pair with the Ace of Clubs is different from the Ace of Clubs chosen first followed by the ace of hearts. This is a double counting.

[quote="InProgress"]Well Luke, what did you use as your denominator?[/quote]
Thanks, InProgress, for answering my inquiry. I used, as math154 did, $\dbinom{52}{3}$ as my denominator.

Wait, I want to do a little casework.

Say you choose two cards of a value $x$ and another card of value $y$. The possible ways you could do this is:
1) Choose the two $x$ cards and then the $y$ card.
2) Choose an $x$ card, the $y$ card, and the other $x$ card.
3) Choose the $y$ card and then the two $x$ cards.

The probability of case 1 is:
$(\frac{1}{13})(\frac{3}{51})(\frac{48}{50})=\frac{1}{11050}$

The probability of case 2 is:
$(\frac{1}{13})(\frac{48}{51})(\frac{3}{50})=\frac{48}{11050}$

The probability of case 3 is:
$(\frac{12}{13})(\frac{4}{51})(\frac{3}{50})=\frac{48}{11050}$

Adding the probabilities up:
$\frac{1}{11050}+\frac{48}{11050}+\frac{48}{11050}=\frac{97}{11050}$

What went wrong here?

Guys, 72/425 is right, I did it in alcumus and it was right :rotfl:  :oops_sign:  :ninja:  :first:

I complementary counted, probably a bit messier than it had to be, but it worked.  :-D

Wait I got a close answer to Luke Skywalker. [hide]6/425[/hide]

But I [hide=accounted for order]$\frac{13\cdot 4\cdot 3 \cdot 12}{52\cdot 51\cdot 50}$ simplifies to 6/425. 13 for the first rank to choose, then 4 cards and then 3 cards of that rank to choose from. And then there's 12 choices for the next rank.hide]. And then I [hide=didn't.]Same numerator. But the denominator was $\binom{52}{3}$, which is 22100. And then it simplified to $\boxed{\frac{36}{325}}$[/hide]. Both ways were incorrect...:( :(

[quote=ilikecake]I just want to make sure of something- when dealing with cards chosen randomly, do I always pay no respect to order? Because in this problem I did and got the wrong answer.[/quote]

ilikecake, I agree. Compare my above work with yours...:(

I'm kinda confused on this
I got both attempts wrong
I did:
denom is 22100. that i get.
for numerator, i did 13C2 for choosing two ranks, then multiplied by 16 for choosing suites..... I got 168/425....What's wrong here?

I solved it by choosing the ranks separately because one rank is used twice whereas the other rank is only used once in the hand. Here is my solution, with steps where you can try solving it on your own.

[hide=Solution]To begin, we will choose the cards that will be the pair. We shall do this by choosing their rank and then suits. *Try doing this yourself before clicking [hide=here]There are $13$ ways to find a rank and $\dbinom42=6$ ways to choose two ranks out of four, making a total of $78$ ways for the first two cards. We next choose the card that completes the hand but [I]cannot be the same rank as the two cards we have just chosen[/I] *Again, try doing this yourself before clicking [hide=here]There are $48$ cards that are not the same rank as our pair. Thus, there are $78\cdot 48 = 3744$ ways to choose a pair out of a standard 52-card deck, for a probability of $\dfrac{3744}{22100}=\boxed{\dfrac{72}{425}}$.[/hide].*[/hide].*[/hide]

Please let me know if you have any questions! :)

Wait this is like a few lines solution.
[hide=Solution]$\binom{3}{2}=3$ ways to choose the pair. Then the probability of $1$ for the first number in the pair, and then $\frac{3}{51}=\frac{1}{17}$ for the second number in the pair. Finally the remaining card has $\frac{48}{50}=\frac{24}{25}$. Multiplying all yields $\frac{72}{425}$.[/hide]

Using complementary counting we get [hide=this] $1-\frac{52\cdot48\cdot44\cdot\frac{1}{6}}{\dbinom{52}3}$ - $\frac{13\cdot\dbinom43}{\dbinom{52}3}=\frac{72}{425}$ [/hide].
PM if you want to know how I got it.

[quote=UnstoppableGoddess]Why can't we do [hide]
$(52*3*48)/(52*51*50)=24/425$
I did 52 because you can choose any card as the first, 3 because it has to be of the same rank as the first but one of the 4 cards with the same rank was taken by the first card, and 48 because the third card can't be of the same rank as the first 2.
There are $52 * 51 * 50$ ways to choose 3 cards without restrictions, and I counted order in both the number of successes and possibilities. So what's wrong? [/hide]?[/quote]

There are two more possibilities. You did $52*3*48$, which corresponds to picking a card, then one with the same rank, then a different rank. You can also pick a card, then a card with a different rank, then one with the same rank as the first which gives $52*48*3$ possibilities again. And finally, you can pick a card, then a card with a different rank, then a card with the same rank as the second, which gives $52*48*3$ possibilities again. This means you have 3 times more possibilities than you thought, which is $72/425$.

[quote=kathyxue]What about just $1*\frac{3}{51}*\frac{48}{50}*6$?
$1$: We just choose a card, it doesn't matter which with a probability of 1
$3/51$: There are 3 cards left of the same suit and 51 cards left in total, so probability is 3/51
$48/50$: There are 48 other cards and 50 cards left so the probability is 48/50
$6$: All cards are distinguishable, so 3!=6
However
$\asy 1*\frac{3}{51}*\frac{48}{50}*6=\frac{144}{425}$[/quote]

You have incorrect latex...

Before I solved this I thought the answer would be much lower.

tysm guys! This helped me understand the solution better.  :thumbup:


Y'know, the AoPSWiki has existed for a long time without copyrights, and doubtlessly will continue to even if none are established. But, I really wish that a documentation license would be decided upon so I would know what I could and couldn't do with the content.

Mr. Crawford said that "A documentation license will be decided upon later." in June 2006. I'm fairly sure it's "later" now.

What do you want to do with the wiki content?

I don't think he wanted to do anything; it's more of a rhetorical question I guess.

See [url]http://www.artofproblemsolving.com/Wiki/index.php/AoPSWiki:General_disclaimer[/url], and in particular paragraph #7 of the [url=http://www.artofproblemsolving.com/Legal/AoPS_L_Comm_Terms.php]AoPS Terms and Conditions[/url].

Beyond what's in this document, what exactly are you looking for?

Thanks; that's what I'm looking for - I didn't think to look for a sitewide license. A few questions: What does "fully-paid-up" mean? What rights exactly does the user retain over the content submitted? #7 addresses AoPS's right to the content, but not the user's. Could the user re-release the content under a GPL license or a copyright?


The plane is divided into strips of width $1$ by parallel lines (a strip - the region between two parallel lines). The points from the interior of each strip are coloured with red or white, such that in each strip only one color is used (the points of a strip are coloured with the same color). The points on the lines are not coloured. Show that there is an equilateral triangle of side-length $100$, with all vertices of the same colour.

put a side that we note $AB$ of the triangle on a line separing a red stripe with a white one (this line trivialy exists), then because the altitude of the triangle is irattional, then the $3$rd vertice of the equilateral triangle must be inside a red or white stripe, therefore we can move a little the side $AB$ , up or down in the respective colour, in which the $3$rd vertice also is, therefore we obtain the desired triangle.