If M and N are Integers and 6^10*4^(2n+1)=2^20*12^M what is

[dlginsberg89]

The problem is from Advantage Testing's GMAT Math Summary Review 3 number 76.

I'm really stumped on this one:

If M and N are Integers and 6^10*4^(2n+1)=2^20*12^M what is the value of n.

a) 7
b)10
c) 12
d) 28
e) 30

There are two unknown variables and some nasty exponents to deal with. First I tried to make the bases 2, but 6 cannot be reduced any further. In the end I got 6^10= 2^(4m-4n-18) but that can't be right because then the value of n would be dependent on the value of m is unknown. Basically I'm stuck... I know this is harder than something that would be on the exam but I'd like to see how somebody would approach it.

[jlucero]

I'd say that this question could be comparable to something you'd see on test day. One issue though- rather than divide parts of these equations by each other (especially with variables), think of this problem as an equation where there are a whole bunch of 2s and 3s on both sides of the equation. Each of 6, 4, and 12 are groups of 2s and 3s and if we can break down the equation given to those prime factors, the equation will be easier to divide (or in this case, eliminate anything that shows up on both sides of the equation).

6^10 * 4^(2n+1) = 2^20 * 12^M
2^10 * 3^10 * 2^2(2n+1) = 2^20 * 2^2(m) * 3^m
2^10 * 3^10 * 2^4n+2 = 2^20 * 2^2m * 3^m

Now notice that we have exactly 10 3s on the left side of this equation, so that means there must be 10 3s on the right side. m=10. Once you plug in m as 10, you can get rid of all the 3s and you're left with:

2^10 * 2^4n+2 = 2^20 * 2^ 20
2^4n+2 = 2^30
4n+2 = 30
n = 7

Takeaway: there's a bunch of 2s and 3s on both sides of the equation. Let's deal with those groups as separate integers and simplify things in terms of those simplest prime factors.

[dlginsberg89]

That my friend was impressive. Thanks so much!

[jnelson0612]

Great! :-)