Problem:
Let S be the sum of all numbers of the form a/b, where a and b are relatively prime positive divisors of 1000. What is the greatest integer that does not exceed S/10?
Solution:
Because 1000=2353. cach a/b may be written in the form 2m5n, where −3≤m≤3 and −3≤n≤3. It follows that each a/b appears exactly once in the expansion of
(2−3+2−2+⋯+22+23)(5−3+5−2+⋯+52+53)
Thus S=2−124−2−3⋅5−154−5−3=8127⋅12519531=2480+1000437, so 10S=248+10000437.
The problems on this page are the property of the MAA's American Mathematics Competitions