1. The problem statement, all variables and given/known data A number divisible by both 16 and 15 is divisible by at least how many unique integers? 2. Relevant equations Combination formula, nCk = n!/(k!(n-k)!) 3. The attempt at a solution I found the prime factorization for 16 and 15, 2x2x2x2x3x5. A number divisible by 16 and 15 must be divisible by any combination of these prime factors. The repeated elements (the four 2's) are giving me trouble. I looked at just the factors 2x3x5 and summed the number of combinations that can be made with them (3C1+3C2+3C3) and got a total of 7. By hand, I wrote out the the number of possible combinations for 2x2x3x5 and got: 2,3,5,2x2,2x3,2x5,3x5,2x2x3,2x2x5,2x3x5,2x2x3x5, which gives 11 total combinations. It looks like if I add another factor of 2 (2x2x2x3x5) then I will gain another 4 possible combinations, and 4 more for every additional "2". So, my solution is that there are 19 possible unique combinations and thus at least 19 unique integers that this number is divisible by. I'm not sure if that's right and I have no real intuition for why 19 (might) be correct. If the problem had asked for a number divisible by both 16 and 45 I'd be really lost (since there would now be repeated 3's as well as repeated 2's)! Is there a general solution to finding the number of combinations with repeated elements??