First of all..let me correct the question...


What is the least value of x, and not the last value of x.

Eliminate 2; for 6^2 = 36; not a multiple of 48

When 6 is in the numerator and 48 in the denominator, the fraction becomes 1/8

6/48 = 1/8. Now for the fraction to be a whole integer it is important that 8 is cancelled. If you multiply it by one more 6, it becomes 6/8 = 3/4.

Here goes the sequence...

6/48=1/8
6*(1/ 8 )=6/8=3/4
6*(3/4) = 9/2
6*(9/2) = 27
Hence the least value of x has to be 4.
I wish I could explain it better, but 4 is definitely the answer

Is there anyway we can do this problem with primes? Break 48 into primes and we know that we need a multiple of 48 with at least 2, 2, 2, 2, 3? Using my MGMAT class lingo i would say 2, 2, 2, 2, 3 are in 48's prime box. And we know that 6^x has 2, 3 as its prime. Now every time we have 6 rasied to a power it will gain the primes of what number its multipled by. So if we raise 6^2 then it now has 2, 2, 3, 3 as primes. So in order to match the # of primes for 48 we must go higher and we know every time we raise by a power we will add a 2, 3. Does that seem right? If my logic is correct i we know we have to raise 6^4 to get the same # of primes as 48.

The answer is 4.

6^x is a multiple of 48, so 6^x/48


2 is wrong

6*6/48 is not a multiple

Try 4

6*6*6*6/48 is 6*6*6*6/6*2*2*2=9

6^4 is a multiple of 48, and the least multiple of the answer choices

Ans=4

6^x= (3x2)^x =(3^x)(2^x)

lets say (3^x)(2^x)=48

==> (3^x)(2^x)=(2^4)(3)

==>x=4

6^x/48

=2^x * 3^x/2^4*3

2^(x-4) * 3(x-1)

To get the least value so that 48 divides 6^x,sub x=4

2^0* 3^3=27

Hence 4