Problem:
Let an=6n+8n. Determine the remainder on dividing a83 by 49.
Solution:
For n odd we may write
an=(7−1)n+(7+1)n=(7n−(1n)7n−1+⋯−1)+(7n+(1n)7n−1+⋯+1)=2(7n+(2n)7n−2+⋯+(n−3n)73+(n−1n)7)=2⋅49(7n−2+(2n)7n−4+⋯+(n−3n)7)+14n
It follows that a83=49k+14⋅83=49k+1162, where k is an integer. Thus, the remainder on dividing a 83 by 49 is the same as the remainder on dividing 1162 by 49. That remainder is 35.
The problems on this page are the property of the MAA's American Mathematics Competitions