HELP

1440000 = 2^8×3^2×5^4 (14 prime factors, 3 distinct)

(8+1) x (2+1) x (4+1) =135 divisors.

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   

$4^3 * 5^4 *6^2=2^6*5^4*2^2*3^2=2^8*3^2*5^4.$ There is actually a neat way to calculate the number of factors a number has. Suppose we have $4^k$ , and the number of factors that number would have is $4^{k+1}$ . Following the same example here, we have $9*3*5=27*5=\boxed{135}$   factors.