Problem:
Positive integers a and b satisfy the condition
log2(log2a(log2b(21000)))=0
Find the sum of all possible values of a+b.
Solution:
The given equation implies that log2a(log2b(21000))=1, then log2b(21000)=2a, and 21000=2b⋅2a. Therefore b⋅2a=1000=125⋅23. The possible solutions for (a,b) are (1,500),(2,250), and (3,125). The requested sum is 501+252+128=881.
The problems on this page are the property of the MAA's American Mathematics Competitions