idontknowwhy94 wrote:

If a, b, x, y are positive integers and \(a^{x}b^y= 144\), then a + b = ?

1. x > y

2. xy = 8

Dear

idontknowwhy94,

I'm happy to respond.

Here, it's helpful to take the

prime factorization of 144:

144 = 12*12 = 3*4*3*4 = (3^2)(4^2) = (3^2)(2^4)

Statement #1: x > y

The exponent of

a must be larger. Certainly this could be (2^4)(3^2), so that x = 4, y = 2, and x > y. Thus, x + y = 6

Remember also, that if b = 12, and y = 2, then

a would have to equal 1, and x could be any power on earth.

We could have (1^137)*(12^2), so that x = 137, y = 2, and x > y. This also works, but x + y = 139, a different answer.

Two different choices, two answer to the prompt question. This statement, alone and by itself, is

insufficient.

Statement #2: xy = 8

This is an interesting restriction.

Case i: we still can use (2^4)(3^2), so that xy = 8. Again, this gives x + y = 6.

Case ii: suppose b = 144. Then y = 1, a = 1, and we could pick x = 8.

144 = (1^8)(144^1), and this also give xy = 8, so this is also valid. This gives x + y = 9.

Once again, two different choices give two answer to the prompt question. This statement, alone and by itself, is also

insufficient.

Combined:

The problem is, the two cases we used for statement #2 also worked for statement #1.

Case i: 144 = (2^4)(3^2)

Case ii: 144 = (1^8)(144^1)

Both of these scenarios satisfy both statements, but they give different sums for x + y. Even with all the information, we cannot determine a unique answer to the prompt question. Even with both statements, everything is

insufficient.

Answer =

(E) Does all this make sense?

Mike

_________________

Mike McGarry

Magoosh Test Prep

Education is not the filling of a pail, but the lighting of a fire. — William Butler Yeats (1865 – 1939)