Find the smallest positive integer having divisors ending in
0, 1, 2, 3, ... 8, 9.
The number must be divisible by 10, dealing with 1,2,5,and 0. The most economical way to churn through divisors ending in odd numbers is to use powers of 3. 2*3*5 gives us 3 and 6. 2*3^2*5 gives us 8 and 9. 2*3^3*5 supplies 4 and 7.
So the answer is 270.

Posted by broll
on 20151203 11:14:52 