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Challenge question: If j and k are positive integers, and (k

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Challenge question: If j and k are positive integers, and (k

by Brent@GMATPrepNow » Sun Jul 01, 2018 7:25 am

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If j and k are positive integers, and (k)(j^6) = (29^29)(11^11), then how many possible values of k are there?
A) 8
B) 10
C) 12
D) 15
E) 18

Answer: B
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by GMATGuruNY » Sun Jul 01, 2018 9:04 am
Brent@GMATPrepNow wrote:If j and k are positive integers, and (k)(j^6) = (29^29)(11^11), then how many possible values of k are there?
A) 8
B) 10
C) 12
D) 15
E) 18
Options for j such that j� will divide into 29²�11¹¹:

j=1
j=11
j=29
j=29²
j=29³
j=29�
j=11*29
j=11*29²
j=11* 29³
j=11*29�

Since there are 10 options for j, there are 10 options for k.

The correct answer is B.
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by Brent@GMATPrepNow » Tue Jul 03, 2018 8:03 am
Brent@GMATPrepNow wrote:If j and k are positive integers, and (k)(j^6) = (29^29)(11^11), then how many possible values of k are there?
A) 8
B) 10
C) 12
D) 15
E) 18
Let's focus on the value of j.

Since j is an integer, it must be the case that j^6 equals some power of 6.
So, for example, j^6 could equal 29^6.
Likewise, j^6 could equal 29^12, because we can rewrite 29^12 as (29^2)^6 in which case, we can see that (29^2)^6 is a power of 6
Likewise, j^6 could equal 29^18, because we can rewrite 29^18 as (29^3)^6 in which case, we can see that (29^3)^6 is a power of 6
etc...

So, if j^6 = (29^x)(11^y), x can equal 0, 6, 12, 18 or 24 (5 different values), and y can equal 0 or 6 (2 different values)
If x can have 5 different values, and y can have 2 different values, then the number of ways to assign values to x and y = (5)(2) = 10
This means j^6 can have 10 different values, which means k can also have 10 different values.

Answer: B

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Brent
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