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# If j and k are positive integers, and

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GRE Instructor
Joined: 10 Apr 2015
Posts: 1432
Followers: 54

Kudos [?]: 1360 [1] , given: 8

If j and k are positive integers, and [#permalink]  01 Jul 2018, 07:35
1
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Question Stats:

50% (01:27) correct 50% (01:01) wrong based on 4 sessions
If j and k are positive integers, and $$kj^6 = (29^{29})(11^{11})$$, then how many possible values of k are there?

A) 8
B) 10
C) 12
D) 15
E) 18
[Reveal] Spoiler: OA

_________________

Brent Hanneson – Creator of greenlighttestprep.com

Intern
Joined: 26 Jun 2018
Posts: 9
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Kudos [?]: 1 [0], given: 0

Re: If j and k are positive integers, and [#permalink]  01 Jul 2018, 12:26
B...10 values for K
GRE Instructor
Joined: 10 Apr 2015
Posts: 1432
Followers: 54

Kudos [?]: 1360 [0], given: 8

Re: If j and k are positive integers, and [#permalink]  03 Jul 2018, 08:07
Expert's post
GreenlightTestPrep wrote:
If j and k are positive integers, and $$kj^6 = (29^{29})(11^{11})$$, 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.

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

Re: If j and k are positive integers, and   [#permalink] 03 Jul 2018, 08:07
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