Find the smallest positive integer n such that n has exactly 144 distinct positive divisors and there are 10 consecutive integers among them (Note: 1 and n are both divisors of n)
I looked at Steve's comment, so that's where I got my ideas.
I start with the smallest 10 consecutive integers possible:
1,2,3,4,5,6,7,8,9,10
The factors required for these divisors:
2^3*3^2*5*7
Now I have to add factors to get 144 divisors. Note that the number of divisors is equal to the product of each exponent plus 1. So there are 4*3*2*2 divisors in the number 2^3*3^2*5*7. Possible factors to add are the following:
2^8 = 256
2^2*5^2 = 100
3*5*7 = 105
2^2*3*5 = 60
3*5*7 appears to lead to Steve's solution. But 2^2*3*5 will improve it. It may possibly be the best solution, but I'm not making any promises.
So, the number is 2^5*3^3*5^2*7 = 151200

Posted by Tristan
on 20060220 12:09:29 