What Is the Last Digit of 1² + 2² + … + 99²? A Classic Math Puzzle

Welcome back to the weekend puzzle section — because what could be better than spending a few minutes on a good, solvable problem? This one once earned a 7th-grader 7 points at a math olympiad in Kaliningrad. Try it yourself before reading the solution.

The Puzzle: Last Digit Sum of Squares from 1² to 99²

Find the last digit of 1² + 2² + … + 99²

Pause here. There is an elegant shortcut hiding in plain sight, and it requires no calculator — only a neat grouping idea.

The Solution: Grouping Into Blocks of Ten

Here is how the last digit sum of squares falls out cleanly:

1² + 2² + … + 99² = (0² + 1² + … + 9²) + (10² + 11² + … + 19²) + … + (90² + 91² + … + 99²)

The key observation: the last digit of n² depends only on the last digit of n. Within each bracket, the last digits of the numbers run through exactly 0, 1, 2, …, 9 — the same set every time. So the sum inside each bracket ends in the same digit, no matter which group of ten you pick.

There are exactly 10 such brackets. Ten copies of the same last digit, added together, always produce a total whose last digit is 0.

The Answer to the Last Digit Problem

The last digit of 1² + 2² + … + 99² is 0.

No need to compute the full sum (which is 328,350, by the way — confirming the answer). The whole argument fits in two lines and works entirely in the language of number theory: periodicity of last digits under squaring.

How Did You Solve It?

This kind of problem — finding a last digit sum of squares through grouping — is a staple of olympiad math precisely because it rewards pattern recognition over brute force. If you got it without the hint, that is a clean win. If the grouping trick surprised you, it is worth keeping in your toolkit: the same idea appears in dozens of last-digit and divisibility problems at every level of competition math.