Problem:
For a positive integer n, let dn be the units digit of 1+2+3+⋯+n. Find the remainder when
n=1∑2017dn
is divided by 1000.
Solution:
Any 20 consecutive positive integers have units digits whose sum is 0+1+2+⋯+ 9+0+1+2+⋯+9=2(1+9)+2(2+8)+2(3+7)+2(4+6)+2(5)=90. Therefore dn+20=dn for all integers n≥1. Thus one only needs to calculate dn for 1≤n≤20, and
(d1,d2,…,d20)=(1,3,6,0,5,1,8,6,5,5,6,8,1,5,0,6,3,1,0,0)
Therefore
n=1∑2017dn=101n=1∑20dn−n=18∑20dn=101⋅70−1=7069
The requested remainder is 69.
The problems on this page are the property of the MAA's American Mathematics Competitions