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Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
If j and k are positive integers, and
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01 Jul 2018, 07:35
01 Jul 2018, 07:35
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38% (02:28) correct
61% (02:27) wrong based on 26 sessions
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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
A) 8
B) 10
C) 12
D) 15
E) 18
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If j and k are positive integers, and
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03 Jul 2018, 08:07
03 Jul 2018, 08:07
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
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
Cheers,
Brent
