2025 AMC 10B Problem 2

The ones digit of $n^2$ is determined by the ones digit of $n$, so Jerry's list repeats with period $10$. Squaring $n = 1, 2, \dots, 10$ gives ones digits
$$1,\ 4,\ 9,\ 6,\ 5,\ 6,\ 9,\ 4,\ 1,\ 0,$$
which sum to $45$.

Now split $2025 = 202 \cdot 10 + 5$. The squares of $1$ through $2020$ form $202$ complete blocks, contributing
$$202 \cdot 45 = 9090.$$
The remaining terms come from $n = 2021, \dots, 2025$, whose ones digits are $1, 2, 3, 4, 5$; their squares end in $1, 4, 9, 6, 5$, a further $25$.

The total is $9090 + 25 = 9115$.

The answer is $\boxed{\textbf{(D)}\ 9115}$.

Only $n \bmod 10$ matters for the ones digit of $n^2$, which turns an apparently enormous sum into one block sum times a count plus a short remainder. The same routine handles any digit-of-a-power sum: find the period, sum one period, divide the range by the period, and add the partial block by hand. The symmetry $(10-k)^2 \equiv k^2 \pmod{10}$ also explains why the block reads as a palindrome around the $5$.

• Taking 2025/10 = 202 complete blocks and stopping at 202 * 45 = 9090, forgetting the five leftover squares, which is choice (C).
• Starting the block at $n = 0$ and summing $0, 1, 4, 9, 6, 5, 6, 9, 4, 1$, which is still $45$ but shifts the leftover terms by one, giving the wrong tail.