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How many natural-number factors does N have if N=2^4*3^8*5^2*7^2?

 Jan 25, 2022
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 N=2^4*3^8*5^2*7^2==128,595,600

 

128595600 = 2^4 * 3^8 * 5^2 * 7^2.  As you can see, it has 4 DISTINCT prime factors{2, 3, 5, 7}

Counting the repeats, then just add the exponents up:4 + 8 + 2 + 2 ==16 prime factors.

 

 

If you want to know the number of "DIVISORS", then it has:

[4+1] * [8+1] *  [2+1] * [2+1] ==5 x 9 x 3 x 3 ==405 DIVISORS.

 Jan 25, 2022

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