Problem:
There is a unique ordered triple (a,k,m) of nonnegative integers such that
2a+2a+k+2a+2k+⋯+2a+mk4a+4a+k+4a+2k+⋯+4a+mk=964.
What is a+k+m?
Answer Choices:
A. 8
B. 9
C. 10
D. 11
E. 12
Solution:
We can use the geometric series formula to write:
2a+2a+k+2a+2k+⋯+2a+mk4a+4a+k+4a+2k+⋯+4a+mk=2a(2k−12(m+1)k−1)4a(4k−14(m+1)k−1)=2a⋅2k+12(m+1)k+1
Note that 2k+12(m+1)k+1 is odd, so a=2, since 964=22⋅241. Hence:
2k+12(m+1)k+1=241
Note that 241=28−24+1, so (m+1)k=12 and k=4. Hence, we have a=2, m=2, and k=4, which yields a final answer of (A) 8.
Remark. Note that the last part can be formalized with Zsigmondy’s Theorem. 2(m+1)k+1 will have a unique prime divisor not dividing 241 or 2k+1 for any (m+1)k>12, hence (m+1)k≤12. It's easy to see 2m+1k≥3⋅241=723, so (m+1)k=10,11,12, of which only k=4,m=2 works.
The problems on this page are the property of the MAA's American Mathematics Competitions