factoring 5 would affect c wouldnt it, because its 5^c...
also, so can u assume that if 2,3 and 5 are taken as factors, then the powers are also divisible by 2, 3 and 5?
Hi, i got to the number being m=2^a * 3^b * 5^c
but how do yout hypthesise these situations:
where (1) b and c are even but a is odd, (2) a and c are divisible by 3 but b leaves a remainder of 1, and (3) a and b are divisible by 5, but c leaves a remainder of 1.
without having worked out the solution being m=30233088000000 yet