Problem:
Let m/n, in lowest terms, be the probability that a randomly chosen positive divisor of 1099 is an integer multiple of 1088. Find m+n.
Solution:
The divisors of 1099 are of the form 2a⋅5b, where a and b are integers with 0≤a≤99 and 0≤b≤99. Since there are 100 choices for both a and b,109 has 100⋅100 positive integer divisors. Of these, the multiples of 1088=288⋅588 must satisfy the inequalities 88≤a≤99 and 88≤b≤99. Thus there are 12 choices for both a and b; i.e., 12⋅12 of the 100⋅100 divisors of 1099 are multiples of 1088. Consequently, the desired probability is nm=100⋅10012⋅12=6259 and m+n=634.
The problems on this page are the property of the MAA's American Mathematics Competitions