How many distinct, natural-number factors does 4^3 * 5^4 *6^2 have?
\(4^3 * 5^4 *6^2\\ =2^6 * 5^4 *2^2*3^2\\ =2^8 * 3^2 *5^4\\\)
2 is the only prime factor, 8
3 is the only prime factor, 2
5 is the only prime factor, 4
2 and 3 are the only prime factors 8*2=16
2 and 5 are the only prime factors 8*4=32
3 and 5 are the only prime factors 2*4=8
2 and 3 and 5 are the only prime factors 8*2*4=64
8+2+4+16+32+8+64 = 134
plus 1 I guess
So that is 135