2017 AIME I Problem 3

Attempt Problem 3 of the 2017 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AIME I solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

3.

For a positive integer n,n, let dnd_n be the units digit of 1+2+3++n.1 + 2 + 3 + \cdots + n. Find the remainder when n=12017dn\sum_{n=1}^{2017} d_n is divided by 1000.1000.

Answer: 69
Concepts:triangular numberunits digitpattern recognition
Difficulty rating: 2300
Solution:

Here dnd_n is the units digit of the triangular number n(n+1)2.\frac{n(n+1)}{2}. Since (n+20)(n+21)2\frac{(n+20)(n+21)}{2} n(n+1)2=20n+210- \frac{n(n+1)}{2} = 20n + 210 is a multiple of 10,10, the sequence dnd_n is periodic with period 20.20. Computing one period, (d1,d2,,d20)=(1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0), \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 70.70.

Since 2017=10020+17,2017 = 100 \cdot 20 + 17, the total is 10070100 \cdot 70 plus the first 1717 terms of the period, which sum to 70(1+0+0)=69.70 - (1 + 0 + 0) = 69. The sum is 7069,7069, so the remainder is 69.69.

← Problem 2#2
Full Exam

Problem 3 in Other Years