Problem explanation

[rastis.johnson]

Can you please do #1 on pg 81 in the newer number properties book? I cannot manipulate the problem to equal that of the solution

[jnelson0612]

Sure. Here's the question:

If x, y, and z are integers, is x even?

1) 10^x = (4^y)(5^z)

2) 3^(x+5) = 27^(y+1)

Let's tackle #1 first by breaking everything into prime factors:

10^x = (2*5)^x which is 2^x * 5^x.

4^y = [(2)^2]^y which is 2^2y. (If you have any trouble with this one, write out 2 squared to the y. See how you want to multiply the square * y to get 2^2y.

Let's put it all together:
2^x * 5^x = 2^2y * 5^z.

What have we learned? x=2y, so x must be even. Sufficient.

Statement 2:
2) 3^(x+5) = 27^(y+1)
Again, let's break it down piece by piece:

27^(y+1) is [(3)^3]^y+1
That turns into 3^(3y+3)

So 3^(x+5) = 3^(3y+3)

Thus x + 5 = 3y + 3

So x + 2 = 3y.

Let's test numbers. x could be 1 and y could be 1. Is x even? NO

OR, x could be 4 and y could be 2. Is x even? YES

I have a YES and a NO: not sufficient!

The answer is A.

[Isabell]

Let's put it all together:
2^x * 5^x = 2^2y * 5^z.

Hi,

could you please explain how you concluded from the equation

2^x * 5^x = 2^2y * 5^z

that x=2y?

Where did the other x and the z disappear? Also, couldn't I divide by 2^2y and 5^x to get: 2^(x-2y)=5^(z-x)? But that would make things a lot more complicated, although it makes more sense to me....

Thanks!

[tim]

because x, y, and z are integers, the powers of 2 must be equal and the powers of 5 must be equal. if you want the complicated reason why, it's because of the unique prime factorization guaranteed by the Fundamental Theorem of Arithmetic. you don't need to know that for the GMAT though, just know that you can set the exponents equal for each of the powers..

your approach works too. once you have your equation, again by the FTA the only integer power of 2 that is equal to an integer power of 5 is 2^0 = 5^0