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