5 cool math tricks ft. Technicality

In some of these tricks, there is multiplication/division by powers of 10. It sounds less imposing to call this, "shifting."
That is, you keep the decimal point (dp) fixed, and shift all the digits right or left some number of places.
Or keep the digits fixed and move the dp that many places in the opposite direction.
To multiply by 10, shift left 1 place/or move the point right 1.
To multiply by 100, shift left 2 places/move the point right 2.
To divide by 10, shift right 1 place/move the point left 1.

1. Taking 20%; i.e., multiplying by 0.2; i.e., dividing by 5:
Move the dp left 1 and double.
BTW, you can do the same, but in reverse, to *multiply* by 5

2. Multiplying by 11:
The trick she gives for two-digit numbers is fine, but what about longer numbers? Do this:
Shift the number left 1 place, and add (to the original number). Of course, this just amounts to the usual method of muliplication. But you can adapt it to mental math by adding consecutive pairs of digits – the complication is applying all the carries. But if you're lucky and there are none, you can multiply by 11 lickety-split!
11 x 21536:
(0)+2=2
2+1=3
1+5=6
5+3=8
3+6=9
6+(0)=6
ANS: 236896

3. Multiplying by 9:
Once again, to extend this to more-than-2-digit numbers, do this:
Shift the number left 1 place (don't forget to slap a trailing "0" onto it!), and subtract the original number.
Her discovery at 3:08, works also in larger bases. You can verify it by treating the muliplication as being on polynomials, where the base you're in ("b"), is the variable:
(base b) (b–1)(b–2) = b² – 3b + 2 = (b–3)b + 2
. . so the base-b digits of the product are b–3 and 2

4. The square root trick, using geometry.
Unfortunately, even confined to just the (positive) integers, this trick will work only when the number can be written as a sum of two squares, n = x² + y², so that you can use x and y as the legs of your right triangle.

Fortunately, though, there are quite a few 2-square numbers. Aside from the squares, up through 50:
2, 5, 8, 10, 13, 17, 18, 20, 26, 29, 32, 34, 37, 40, 41, 45, 50
.. which are subject to the Pythagorean Theorem trick.
The non-2-square numbers are:
3, 6, 7, 11, 12, 14, 15, 19, 21, 22, 23, 24, 27, 28, 30, 31, 33, 35, 38, 39, 42, 43, 44, 46, 47, 48
.. and square roots of these can't be found by that method, as shown. An extension of it, however, will work for any (positive) integer.
[Hint: Every positive integer can be expressed as a sum of 4 or fewer squares.]

5. Chinese Remainder Theorem.
Pretty neat! Look it up for more info. It applies to any finite number of co-prime integers; it gets more complicated the more of them there are.


ADDENDUM:
4. When the number you want the √ of, isn't a sum of two squares (or even when it is!), you can always write it as a *difference* of two squares (even though sometimes those will have to be squares of half-integers), using the difference-of-squares factorization:
x² – y² = (x – y)(x + y)
For instance, 7 can't be written as a sum of fewer than 4 squares:
7 = 2² + 1² + 1² + 1²
but every odd integer is a difference of consecutive squares:
2n + 1 = (n+1)² – n²   ;    with n=3:
7 = 4² – 3²

So measure *c* = 4 units from one end of your pipe cleaner;
bend it there and measure *b* = 3 units;
bend it there into a 90º angle. The 3rd segment will be *a* when we're done.
Now, maintaining the 90º angle between *b* & *a*, flex the angles between *c* and *b* until the 3rd segment closes the triangle (meets the beginning end of the pipe cleaner.) *a* now = √7.

For n = 6, its minimum sum of squares is
6 = 2² + 1² + 1²

But n = 6 = 2·3 = (x – y)(x + y),   x = ½(3+2) = ⁵/₂ , y = ½(3–2) = ½
6 = (⁵/₂)² – (½)²

So do what you did in the previous example, but now with
*c* = 2½
*b* = ½
You will get
*a* = √6