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If j and k are positive integers, and kj^6 = (29^29)

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If j and k are positive integers, and kj^6 = (29^29)  [#permalink]

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New post 01 Jul 2018, 08:34
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Question Stats:

31% (01:09) correct 69% (02:04) wrong based on 36 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

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Re: If j and k are positive integers, and kj^6 = (29^29)  [#permalink]

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New post 01 Jul 2018, 09:20
1
Values of K
29^29 * 11^11
29^29 * 11^5
29^23 * 11^11/ 11^5
29^17 * 11^11/11^5
29^11 * 11^11/11^5
29^5 * 11^11/11^5

Total = 10 values
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If j and k are positive integers, and kj^6 = (29^29)  [#permalink]

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New post 01 Jul 2018, 10:02
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GMATPrepNow 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


Options for \(j\) such that \(j^6\) will divide into \(29^{29}11^{11}\):

\(j=1\)
\(j=11\)
\(j=29\)
\(j=29^2\)
\(j=29^3\)
\(j=29^4\)
\(j=11*29\)
\(j=11*29^2\)
\(j=11* 29^3\)
\(j=11*29^4\)

Since there are 10 options for \(j\), there are 10 options for \(k\).


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Re: If j and k are positive integers, and kj^6 = (29^29)  [#permalink]

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New post 01 Jul 2018, 11:53
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2
GMATPrepNow 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


Given \(kj^6 = (29^{29})(11^{11})\)

We will try to find the possible values for \(j\), since we know its power is a multiple of \(6\)

We have \(4\) powers of \(29\) as multiples of \(6\) = \(29^{6}, 29^{12}, 29^{18}, 29^{24}\)

& \(1\) power of \(11\), as a multiple of \(6\) = \(11^6\)

Hence we have \(4\) ways of choosing \(29\), to substitute for \(j\)

We have \(1\) way of \(11\) to substitute for \(j\)

combination of \(11\) & \(29\), will result in \(4*1 = 4\) more ways to substitute for \(j\)

lastly \(j\) can be \(1\), hence \(1\) more value for \(j\).

So we have \(4 + 1 + 4 + 1 = 10\) values to substitute for \(j\) & hence we have \(10\) values for \(k\).

for e.g.

\(kj^6 = (29^{23})(11^{5})(29^{6})\)

\(kj^6 = (29^{29})(11^{5})(11^{6})\)

\(kj^6 = (29^{17})(11^{5})(11*29^{2})^{6}\)


Answer B.


Thanks,
GyM
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Re: If j and k are positive integers, and kj^6 = (29^29)  [#permalink]

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New post 03 Jul 2018, 09:08
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GMATPrepNow 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.

Answer: B

Cheers,
Brent
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Re: If j and k are positive integers, and kj^6 = (29^29) &nbs [#permalink] 03 Jul 2018, 09:08
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