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?
If N=a^p.b^q.c^r..., (a,b,c... are primes) then it has (1+p)(1+q)(1+r)... divisors.
Here, the possibilities are N=a^15, a^7.b^1, a^3.b^3, a^3.b^1.c^1, a^1.b^1.c^1.d^1. Given that we want the minimum N, we should take a=2, b=3, c=5, d=7 and the possibilities are 32868, 128x3=384, 8x27=216, 8x3x5=120, 2x3x5x7=210... the optimum is 120.
Edited on February 3, 2005, 3:05 pm
Answer
