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2k is a perfect square, 3k is a perfect cube, and 5k is a perfect 5th power. What is the sum of the exponents in the prime factorization of the smallest such positive integer k ?

 Aug 6, 2018
 #4
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2k = 2 x 2 = 4 perfect square. k=2^1 - the exponent =1

 

3k = 3 x 9 =27 perfect cube. k=3^2 - the exponent = 2

 

5k =5 x 625 =3,125 perfect 5th power. k= 5^4 - the exponent =4

Sum of the exponents of k =1 + 2 + 4 = 7

 Aug 6, 2018
 #6
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You misunderstood the question, Synth got it right

 

EDIT: link to the site synth took the answer from (The latex there is readable)

 

link

Guest Aug 6, 2018
edited by Guest  Aug 6, 2018

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