PS question- multiples

[jamesk486]

when 6^x is a multiple of 48, what's the last value of x?
A. 2
B. 4
C. 6
D. 8
E.10

i think this may have to do with primes?

[Cybermusings]

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

[scoutkb]

I posted this in another thread that asked the same quesiton.

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.

[rajeshvellanki]

check https://www.beatthegmat.com/viewtopic.php?t=2307