2017 AIME I Problems/Problem 3

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Problem 3

For a positive integer $n$, let $d_n$ be the units digit of $1 + 2 + \dots + n$. Find the remainder when \[\sum_{n=1}^{2017} d_n\]is divided by $1000$.

Solution

We see that $d(n)$ appears in cycles of $20$, adding a total of $70$ each cycle. Since $\lfloor\frac{2017}{20}\rfloor=100$, we know that by $2017$, there have been $100$ cycles, or $7000$ has been added. This can be discarded, as we're just looking for the last three digits. Adding up the first $17$ of the cycle of $20$, we get that the answer is $\boxed{069}$.

See Also

2017 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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