# Difference between revisions of "2017 AIME I Problems/Problem 3"

## 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 $20$ and the cycles are $$1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0,$$ adding a total of $70$ each cycle. Since $\left\lfloor\frac{2017}{20}\right\rfloor=100$, we know that by $2017$, there have been $100$ cycles and $7000$ 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 $\boxed{069}$.

## Video Solution

https://youtu.be/BiiKzctXDJg ~Shreyas S

## See Also

 2017 AIME I (Problems • Answer Key • Resources) Preceded byProblem 2 Followed byProblem 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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