Problem:
Define Ln as the least common multiple of all the integers from 1 to n inclusive. There is a unique integer h such that
11+21+31+⋯+171=L17h
What is the remainder when h is divided by 17?
Answer Choices:
A. 1
B. 3
C. 5
D. 7
E. 9
Solution:
Note that
1+21+⋯+161=L16m
for some integer m, so the given sum is
L16m+171=L17h
Now L17=17L16, so it follows that h=17m+L16. Therefore h≡L16(mod17). Note that L16=24⋅32⋅5⋅7⋅11⋅13. These factors modulo 17 satisfy 24≡−1,5⋅7≡1,9⋅11≡−3, and 13≡−4, so the product is congruent to (−1)(1)(−3)(−4)=−12≡(C)5(mod17).
The problems on this page are the property of the MAA's American Mathematics Competitions