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805+ (Hard)|   Algebra|   Exponents|      
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idontknowwhy94
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If a, b, x, y are positive integers and a^xb^y= 144, then a + b = ? 05 Oct 2016, 05:34
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95% (hard)

Question Stats:

24% (02:10) correct 76% (02:17) wrong based on 405 sessions

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

1. x > y

2. xy = 8
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E
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mikemcgarry
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If a, b, x, y are positive integers and a^xb^y= 144, then a + b = ? 05 Oct 2016, 10:45
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If a, b, x, y are positive integers and \(a^{x}b^y= 144\), then a + b = ?

1. x > y

2. xy = 8
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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 :-)
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If a, b, x, y are positive integers and a^xb^y= 144, then a + b = ? 12 Aug 2020, 03:00
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idontknowwhy94


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.

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Since it says a,b,x,y are (different?) integers and in cases i and ii, the numbers 2 and 1 appear multiple times. Is that reason enough to conclude that both statements are not sufficient? Thank you!
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If a, b, x, y are positive integers and a^xb^y= 144, then a + b = ? 12 Aug 2020, 06:59
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Since it says a,b,x,y are (different?) integers and in cases i and ii, the numbers 2 and 1 appear multiple times. Is that reason enough to conclude that both statements are not sufficient? Thank you!
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The question does not say that a, b, x and y are different. Two different letters in algebra can represent the same number -- if you see the equation x + y = 2, for example, it is certainly possible that x and y both equal 1.

If you did think the unknowns needed to be different here, you'd find that no solution is even possible. That can absolutely never happen in a real GMAT DS question, because then it's not clear what answer choice is correct (is the information sufficient, because you know with certainty no solution exists, or is it insufficient because there is no solution? There's no way to decide). So if you did make an assumption when solving a DS question, and found no solutions were possible, that would mean your assumption was flawed, and you'd need to correct that assumption and solve the problem again.
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If a, b, x, y are positive integers and a^xb^y= 144, then a + b = ? 12 Aug 2020, 09:40
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If a, b, x, y are positive integers and \(a^{x}b^y= 144\), then a + b = ?

1. x > y --> insuff: \(a^{x}b^y= 1^2*144^1=1^3*12^2=2^4*3^2\), so different values of x & y (x>y), we can get different values of a+b

2. xy = 8--> insuff: \(a^{x}b^y= 1^8*144^1=1^4*12^2=2^4*3^2\), so different values of x & y (xy=8), we can get different values of a+b

Combining (1) & (2) =>
\(a^{x}b^y= 1^8*144^1=1^4*12^2=2^4*3^2\) --> insuff: different values of x & y (xy=8 & x>y), we can get different values of a+b
Answer: E
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If a, b, x, y are positive integers and a^xb^y= 144, then a + b = ? 29 Nov 2022, 10:57
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