Squaring double-digit numbers mentally
Numbers ending in 0
The easiest double-digit numbers to square are those that end in 0, which only requires squaring a single-digit number then appending two zeros to the end of the result. For example, 602 = 6×6×100 = 3600.
Numbers ending in 5
To square a number ending in 5 (not just a double-digit number but any number), there's a quick trick: drop the 5, multiply the remaining number by its successor, then append 25 to the end. For example, 752 = 7×8×100 + 25 = 5625. I'll explain why this works in the last section.
Numbers ending in 1, 4, 6, or 9
Since double-digit numbers ending in 0 or 5 are the easiest to square, the next easiest are the ones immediately before or after those numbers, that is, numbers ending in 9 (before a 0), 1 (after a 0), 4 (before a 5), or 6 (after a 5). To square those numbers, let's see the difference between any two consecutive squares. The difference between a square and the square before it is:
Thus, to square a number ending in 9 or 4, square the easy-to-square number that follows it, then subtract double that easy-to-square number and add one. For example, 592 = 602 − 2×60 + 1 = 3600 − 120 + 1 = 3481.
Similarly, the difference between a square and the square after it is:
so to square a number ending in 1 or 6, square the easy-to-square number before it, add double the easy-to-square number, and add one. For example, 862 = 852 + 85×2 + 1 = 7225 + 170 + 1 = 7396.
This technique can be used with any number, but it's simplest with 1s, 4s, 6s, and 9s because of how easy it is to both square and double the adjacent number ending in 0 or 5.
Numbers ending in 2, 3, 7, or 8
For the remaining numbers, those ending in 2, 3, 7, and 8, we can use this relationship:
I usually pick a such that either x − a or x + a is a multiple of 10, which means it's only ever 2 or 3 and a2 is only ever 4 or 9. For example, 672 = (67 − 3)(67 + 3) + 32 = 64×70 + 9 = 4200 + 280 + 9 = 4489.
This algebraic relationship is the reason that the above trick for squaring numbers ending in 5 works. When x ends in 5, x2 = (x − 5)(x + 5) + 52. That is, multiply the multiple of ten before the number by the multiple of ten after it and add 25.