For this problem, only positive whole numbers are to be considered. What is the smallest number that has exactly 16 numbers (including 1 and itself) that divide into it without remainder? Proof?
The number of proper divisors a number has is equal to the product of one more than each exponent in the prime factorization.
For example 90 has 12 proper divisors because 90=2^1*3^2*5^1 and 2*3*2=12
For a number to have 16 divisors, we need the product to be 16. There are 5 ways to do this:
16, 8*2, 4*4, 4*2*2, 2*2*2*2
For each of these, the smallest prime factorizations are:
2^15=32768
2^7*3=384
2^3*3^3=216
2^3*3*5=120
2*3*5*7=210
The smallest of these is 210 so it is the soution. Its actual factors are a simple matter of combinatorics.
Jer

Posted by Jer
on 20050203 17:09:06 