1440000 = 2^8×3^2×5^4 (14 prime factors, 3 distinct)
(8+1) x (2+1) x (4+1) =135 divisors.
+118608 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
+4609 \(4^3 * 5^4 *6^2=2^6*5^4*2^2*3^2=2^8*3^2*5^4.\) 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.
