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

Question

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

GMATGuruNY · Accepted Answer

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 .

B

pandeyashwin · Answer

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

GyMrAT · Answer

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

BrentGMATPrepNow · Answer

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

rvgmat12 · Answer

Total number of factors = (29+1)*(11+1) = 30*12

Now j^6 has (30/6)*(12/6) = 5*2 = 10 options

so k also has 10 different options

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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)

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