2025 AMC 10B Problem 2

1. Write out the first ten. Square $1$ through $10$ and keep only the ones digit:

These add up to $1+4+9+6+5+6+9+4+1+0 = 45$.

1. Why the list repeats. When you multiply, the ones digit of the answer comes only from the ones digits of the numbers. For example, $13^2 = 169$ ends in $9$, just like $3^2 = 9$. So squares of $11$ to $20$ end in the same ten digits again, and so do $21$ to $30$, and so on. Every block of ten adds to $45$.

3. Count the full blocks. $2025 = 202 \times 10 + 5$. So $1$ to $2020$ makes $202$ full blocks, worth
$$202 \times 45 = 9090.$$

1. Add the leftovers. The last five numbers are $2021, 2022, 2023, 2024, 2025$. They end in $1, 2, 3, 4, 5$, so their squares end in $1, 4, 9, 6, 5$, which add to $25$.

1. The total is $9090 + 25 = 9115$.

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

A last digit never cares about the digits in front of it, so a sum of last digits over a huge range is really one short block repeated. The recipe: find how often the pattern repeats, add one block, multiply by the number of full blocks, and finish the partial block by hand. The partial block at the end is the part people forget.

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 $\boxed{\textbf{(D)}\ 9115}$.