# Quick Math

Here are some tricks I use to calculate things quickly, either in my head or on paper.

## Squaring

Since single-digit multiplication is relatively easy for most people, we can build on that instead of using the traditional method of multiplication on paper and having to juggle a bunch of numbers in our head.

By separating the number into a sum of two smaller numbers (one for the tens place, *x*, and another for the ones place, *y*), we can then use this equation to know how to calculate the result:

${(10x+y)}^{2}=100{x}^{2}+20xy+{y}^{2}$

Broken down into simple instructions, what we do is:

- Square the digit in the tens place.
- Put two zeros at the end of the result.
- Remember that.
- Multiply the tens and ones place.
- Double it.
- Put a zero at the end of the result.
- Add it to the last number you remembered.
- Remember that.
- Square the digit in the ones place.
- Add it to the last number you remembered.

### Example

As an example, say I want to square 29. In the traditional way, first I imagine the number on top of itself and try to multiply 29 by 9. Once I get the result (261) and memorize it, I move on to multiplying 29 by 20, and getting 580. After remembering *that* too, I then finally add those numbers together to get 841.

Now consider doing it the way I mentioned about: I separate 29 in my head into (20 + 9)^{2}. When expanding, this becomes (20 + 9)(20 + 9) and we FOIL; 20 × 20 is 400, 20 × 9 is 180, doubling it gets us 360, add it to 400 and get 760, then finally 9 × 9 is 81 and add it all to get 841.

## Neighboring Square

If you ~for some reason~ need to calculate a neighboring square, here are the equations for reusing your answer and save you some time.

For the neighbor above,

${(x+1)}^{2}={x}^{2}+2x+1$

- Get the square you had before.
- Add the number you were originally given.
- Do it again.
- Add 1.

For the neighbor below,

${(x-1)}^{2}={x}^{2}-2x+1$

- Get the square you had before.
- Subtract the number you were originally given.
- Do it again.
- Add 1.

## Rectangling

We can generalize the trick in the previous section to also make it easier to multiply any two double-digit numbers.

Until I can get around to completing this, it’s left as an exercise to the reader.

## Square rooting

This one I *definitely* should’ve figured out way back when I was taking my physics exams without a graphing calculator. Intentionally.

You can certainly get 2 significant digits in your head with this, although 3 or more may require paper.

The algorithm goes something like this:

- If the given number has an odd number of digits, put a zero at the beginning of it.
- Divide the number of digits by 2. That’s how many digits your answer will have.
- Set the most significant digit of your answer to 1 and the rest to 0.
- We’ll call this the lower square.

- Increment the most significant digit and get its square.
- We’ll call this the higher square.

- If the given number falls within the lower and higher squares, continue. Otherwise, go to step 4.
- Get the difference between the higher and lower squares.
- Divide it by 10.
- Get the difference between the given number and the lower square.
- Estimate how many times the result in (6) can go in the result in (7).
- If you’re split on which digit to use, go with the lower one.

- This is the next digit in your answer.
- Check your answer.
- For each consecutive digit (and virtual bonus points you can spend in your mind palace), go to step 6.