If \(a, b\), and \(c\) are integers, what is the value of

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If \(a, b\), and \(c\) are integers, what is the value of \(a\)?

1) \(2^a+2^b=33\)
2) \(a\cdot c = 5\)

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by ceilidh.erickson » Sat Jun 01, 2019 10:38 am

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If \(a, b\), and \(c\) are integers, what is the value of \(a\)?

We're given no information in the question stem except that these variables are integers. So, we have to dive into the statements:

(1) \(2^a+2^b=33\)
Think of combinations of powers of 2 that would add to 33. Since 33 is odd, it must be (odd + even) or (even + odd). The only power of 2 that's odd is \(2^0=1\) .
\(2^0+2^5=1+32=33\)
We know that one of these values must be 0 and the other 5, but we don't know which is which. Insufficient.

(2) \(a\cdot c = 5\)
If both of these are integers, it must be 1*5 or 5*1. Since we don't know which is which, though, this is insufficient.

(1) and (2) together:
(1) tells us that \(a=0\) or \(a=5\), and (2) tells us that \(a=1\) or \(a=5\). Using the statements together, it must be the case that \(a=5\). Sufficient.

The answer is C.
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by Ian Stewart » Sat Jun 01, 2019 12:43 pm

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ceilidh.erickson wrote: (2) \(a\cdot c = 5\)
If both of these are integers, it must be 1*5 or 5*1. Since we don't know which is which, though, this is insufficient.
Just because this is so important in so many questions: if ac = 5, and a and c are integers, there are four possibilities, not two: a and c can be 5 and 1, in either order, or they can be -5 and -1, in either order.

Of course, when we combine the two statements, we can discard the negative solutions, but from Statement 2 alone, we have four possible values of a.
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by ceilidh.erickson » Sat Jun 01, 2019 5:06 pm

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Ian Stewart wrote:
ceilidh.erickson wrote: (2) \(a\cdot c = 5\)
If both of these are integers, it must be 1*5 or 5*1. Since we don't know which is which, though, this is insufficient.
Just because this is so important in so many questions: if ac = 5, and a and c are integers, there are four possibilities, not two: a and c can be 5 and 1, in either order, or they can be -5 and -1, in either order.

Of course, when we combine the two statements, we can discard the negative solutions, but from Statement 2 alone, we have four possible values of a.
Ian has an excellent point here! The question doesn't specify non-negative. I think this shows how we all really synthesize before we extrapolate - after reading statement 1, when I read statement 2 I immediately thought "well a=5, but I don't know that from this one alone" and didn't further pick apart what I already knew to be insufficient. But Ian's point is important - we can't assume a non-negative constraint where none is specified.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education