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

6Likewise, 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

6etc...

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

_________________

Brent Hanneson – Creator of greenlighttestprep.com

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