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If a and b are positive integers..

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If a and b are positive integers..

by ssyohee » Tue Jan 19, 2016 11:53 pm
If a and b are positive integers and a^(b+4) = a^(2b-4), what is the value of a+b?

(1) a = 3
(2) b-a = 5

I can get b in the equation above and then I know a from (1), so sufficient. I also can get a from (2). so sufficient. I think D. each statement alone is sufficient. but the answer is A. Can someone explain this to me? :o
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by GMATinsight » Wed Jan 20, 2016 6:19 am
ssyohee wrote:If a and b are positive integers and a^(b+4) = a^(2b-4), what is the value of a+b?

(1) a = 3
(2) b-a = 5

I can get b in the equation above and then I know a from (1), so sufficient. I also can get a from (2). so sufficient. I think D. each statement alone is sufficient. but the answer is A. Can someone explain this to me? :o
a^(b+4) = a^(2b-4)
i.e. Either a = 1 AND/OR b+4 = 2b-4
i.e. Either a = 1 AND/OR b = 8

Question : a+b = ?

Statement 1: a = 3
i.e. b = 8 for sure
i.e. a+b = 3+8 = 11
SUFFICIENT

Statement 2: b-a = 5
i.e. if b = 8 then a = 3 i.e. a+b = 11
and if a = 1 then b = 6 i.e. a+b = 7
NOT SUFFICIENT

Answer: option A
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by ceilidh.erickson » Thu Jan 21, 2016 8:32 am
Beware of the GMAT question that seems too obvious!

Here, you might think that we can solve the original equation for b. If a and b are positive integers and a^(b+4) = a^(2b-4), then b + 4 = 2b - 4, right? Take a moment to question that. Are there any other possibilities?

Usually, if we set the bases equal, we can set the exponents equal. If 3^x = 3^(y - 1), then x = y - 1, because every power of 3 is unique. There are 3 exceptions to this rule, though: 0, 1, and -1.
1 raised to any power = 1, so we can't set the powers equal
0 raised to any power = 0, so we can't set the powers equal
-1 will alternate between positive 1 and -1, based on whether the exponent is even or odd, but the principle is the same.

So in your problem, one of two things must be true:
either a > 1 and therefore b + 4 = 2b - 4
or... a = 1

If you consider both of these possibilities, you can see that only statement 1 eliminates the 2nd scenario, and leave us with definite values for a and b.

Answer: A
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by Matt@VeritasPrep » Fri Jan 22, 2016 3:54 pm
Since a and b are positive, aᵇ�� = a²ᵇ�� implies one of two things:

Thing #1: a = 1, b = any positive integer.
Thing #2: a ≠ 1, b = 8.

S1 tells us that a = 3, so Thing #2 is true, and b = 8; SUFFICIENT.

S2 tells us that (b - a) = 5. If a = 1, then b = 6. If a ≠ 1, then b = 8 and a = 3. So we have two possible solutions, for a (very tricky!!) NOT SUFFICIENT.
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