2017 AIME I 第 3 题

$d_n$ is the units digit of $\frac{n(n+1)}{2};$ look for a repeating pattern in these digits

The digits repeat with period $20,$ and one full period of $20$ values sums to $70$

Here $d_n$ is the units digit of the triangular number $\frac{n(n+1)}{2}.$ Since $\frac{(n+20)(n+21)}{2}$ $- \frac{n(n+1)}{2} = 20n + 210$ is a multiple of $10,$ the sequence $d_n$ is periodic with period $20.$ Computing one period, $$\begin{aligned} &(d_1, d_2, \ldots, d_{20}) \\ &\quad \tiny = (1, 3, 6, 0, 5, 1, 8, 6, 5, 5, 6, 8, 1, 5, 0, 6, 3, 1, 0, 0), \end{aligned}$$ which sums to $70.$

Since $2017 = 100 \cdot 20 + 17,$ the total is $100 \cdot 70$ plus the first $17$ terms of the period, which sum to $70 - (1 + 0 + 0) = 69.$ The sum is $7069,$ so the remainder is $69.$