+0

plz help fast urgent

+2
153
6
+266

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

#1
+511
+12

This is what I found:

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.

Aug 6, 2018
#2
+266
+2

can you ecplain that modulo part

Darkside  Aug 6, 2018
#3
+511
+11

I don’t know, sorry

Aug 6, 2018
#4
+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

Aug 6, 2018
#6
+1

You misunderstood the question, Synth got it right

Guest Aug 6, 2018
edited by Guest  Aug 6, 2018
#5
+511
+10

Thx for help

Aug 6, 2018