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
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Brent Hanneson - founder of Greenlight Test PrepSign up for our GRE Question of the Day emails