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

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Joined: 12 Sep 2015
Posts: 2864
If j and k are positive integers, and kj^6 = (29^29)  [#permalink]

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01 Jul 2018, 08:34
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4
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Difficulty:

75% (hard)

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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Brent Hanneson – GMATPrepNow.com

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Joined: 14 Jun 2018
Posts: 62
Re: If j and k are positive integers, and kj^6 = (29^29)  [#permalink]

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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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Joined: 04 Aug 2010
Posts: 273
Schools: Dartmouth College
If j and k are positive integers, and kj^6 = (29^29)  [#permalink]

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01 Jul 2018, 10:02
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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

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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01 Jul 2018, 11:53
1
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.

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

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

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

Thanks,
GyM
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Posts: 2864
Re: If j and k are positive integers, and kj^6 = (29^29)  [#permalink]

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03 Jul 2018, 09:08
Top Contributor
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.

Cheers,
Brent
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Brent Hanneson – GMATPrepNow.com

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