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16 Sep 2014, 01:53
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B
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Difficulty:
95%
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Question Stats:
33% (01:58) correct
67%
(02:13)
wrong
based on 90
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Positive integers \(a\), \(b\), \(c\), \(m\), \(n\), and \(p\) are defined as follows: \(m = 2^a3^b\), \(n = 2^c\), and \(p = \frac{2m}{n}\). Is \(p\) odd?
(1) \(a \lt b\)
(2) \(a \lt c\)
(1) \(a \lt b\)
(2) \(a \lt c\)
16 Sep 2014, 01:53
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Official Solution:
Positive integers \(a\), \(b\), \(c\), \(m\), \(n\), and \(p\) are defined as follows: \(m = 2^a3^b\), \(n = 2^c\), and \(p = \frac{2m}{n}\). Is \(p\) odd?
We should first combine the expressions for \(m\), \(n\), and \(p\) to get the following:
\(p = \frac{2m}{n} = \frac{2(2^a3^b)}{2^c} = 2^{a + 1 - c}3^b\)
The question can be rephrased as "Does \(p\) have no 2's in its prime factorization?" Since \(p\) is an integer, we know that the power of 2 in the expression for \(p\) above cannot be less than zero (otherwise, \(p\) would be a fraction). So we can focus on the exponent of 2 in the expression for \(p\): "Is \(a + 1 - c = 0\)?" In other words, "Is \(a + 1 = c\)?"
Statement (1): INSUFFICIENT. The given inequality does not contain any information about \(c\).
Statement (2): SUFFICIENT. We are told that \(a\) is less than \(c\). We also know that \(a\) and \(c\) are both integers (given) and that \(a + 1 - c\) cannot be less than zero.
In other words, \(a + 1\) cannot be less than \(c\), so \(a + 1\) is greater than or equal to \(c\). The only way for \(a\) to be less than \(c\) AND for \(a + 1\) to be greater than or equal to \(c\), given that both variables are integers, is for \(a + 1\) to equal \(c\). No other possibility works. Therefore, we have answered our rephrased question "Yes."
Answer: B
Positive integers \(a\), \(b\), \(c\), \(m\), \(n\), and \(p\) are defined as follows: \(m = 2^a3^b\), \(n = 2^c\), and \(p = \frac{2m}{n}\). Is \(p\) odd?
We should first combine the expressions for \(m\), \(n\), and \(p\) to get the following:
\(p = \frac{2m}{n} = \frac{2(2^a3^b)}{2^c} = 2^{a + 1 - c}3^b\)
The question can be rephrased as "Does \(p\) have no 2's in its prime factorization?" Since \(p\) is an integer, we know that the power of 2 in the expression for \(p\) above cannot be less than zero (otherwise, \(p\) would be a fraction). So we can focus on the exponent of 2 in the expression for \(p\): "Is \(a + 1 - c = 0\)?" In other words, "Is \(a + 1 = c\)?"
Statement (1): INSUFFICIENT. The given inequality does not contain any information about \(c\).
Statement (2): SUFFICIENT. We are told that \(a\) is less than \(c\). We also know that \(a\) and \(c\) are both integers (given) and that \(a + 1 - c\) cannot be less than zero.
In other words, \(a + 1\) cannot be less than \(c\), so \(a + 1\) is greater than or equal to \(c\). The only way for \(a\) to be less than \(c\) AND for \(a + 1\) to be greater than or equal to \(c\), given that both variables are integers, is for \(a + 1\) to equal \(c\). No other possibility works. Therefore, we have answered our rephrased question "Yes."
Answer: B
12 Oct 2014, 07:12
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Bunuel
Bunuel,
I am not getting your combination in step one. I reviewed exponent properties here https://gmatclub.com/forum/math-number-theory-88376.html but I'm still just not seeing it.
