All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Sweet Sixteener (Posted on 2005-02-03)
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?

 See The Solution Submitted by Richard Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 2 of 5 |

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 2005-02-03 17:09:06

 Search: Search body:
Forums (0)