+0  
 
+2
115
6
avatar+262 

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 ?

Darkside  Aug 6, 2018
 #1
avatar+913 
+12

This is what I found: 

 

Answer:

59

Explanation:

The prime factorization of k is :
k=pn11⋅pn22⋅...⋅pnmm
That factorization must contain 2, 3 and 5 as prime numbers :
⇒k=2n1⋅3n2⋅5n3
n1 must be odd, multiple of 5, and multiple of 3⇒15
n2 must be 2 modulo 3, even, and multiple of 5⇒20
n3 must be 4 modulo 5, even, and multiple of 3⇒24
So, 15+20+24=59 is the number asked.

Synth  Aug 6, 2018
 #2
avatar+262 
+2

can you ecplain that modulo part

Darkside  Aug 6, 2018
 #3
avatar+913 
+11

I don’t know, sorry

Synth  Aug 6, 2018
 #4
avatar
+2

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

Guest Aug 6, 2018
 #6
avatar
+1

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
 #5
avatar+913 
+10

Thx for help

Synth  Aug 6, 2018

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