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How many distinct, natural-number factors does 4^3 * 5^4 *6^2 have?

 Oct 19, 2018
 #1
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+2

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

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

 Oct 19, 2018
 #2
avatar+95000 
+2

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   

 Oct 19, 2018
edited by Melody  Oct 19, 2018
 #3
avatar+3726 
+2

\(4^3 * 5^4 *6^2=2^6*5^4*2^2*3^2=2^8*3^2*5^4.\) 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.

 Oct 20, 2018
 #4
avatar+95000 
+1

Thanks Tertre

Melody  Oct 20, 2018

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