Problem:
Let N=1002+992−982−972+962+⋯+42+32−22−12, where the additions and subtractions alternate in pairs. Find the remainder when N is divided by 1000.
Solution:
Reordering the sum shows that
N=(1002−982)+(992−972)+(962−942)+⋯+(42−22)+(32−12)
which equals
2⋅198+2⋅196+2⋅190+2⋅188+⋯+2⋅6+2⋅4=4(99+98+95+94+⋯+3+2)=4(99+95+91+⋯+3)+4(98+94+90+⋯+2)=4[225(99+3)]+4[225(98+2)]=100⋅(51+50)=10100,
and the required remainder is 100.
The problems on this page are the property of the MAA's American Mathematics Competitions