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If $n = 2^{10} \cdot 3^{14} \cdot 5^{8}$, how many of the natural-number factors of $n$ are multiples of 150

 Apr 6, 2020
 #1
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\(n = 2^{10} \cdot 3^{14} \cdot 5^{8}\)

 

150=5^2*3*2

We have \(2^9*3^{13}*5^6\) for each number of exponent, it can be up to that or less. or none. so we get 10*14*7 soo my answer is 980.

 

i think

 

smileysmileysmiley

 Apr 6, 2020
 #2
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2^10 x 3^14 x 5^8 =

 

2^1 x 3^1 x 5^2 =150

 

Subtract the exponents of the same base numbers and + 1 in each case as follows:

 

[10 - 1 + 1] x [14 - 1 + 1] x [8 - 2 + 1] =10 x 14 x 7 = 980 such numbers.

 Apr 6, 2020

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