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Positive integers a, b, c, m, n, and p are defined as follows: m = 2^a [#permalink]
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19 Nov 2014, 07:22
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Tough and Tricky questions: Exponents. 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) \(a \lt c\) Kudos for a correct solution.
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Re: Positive integers a, b, c, m, n, and p are defined as follows: m = 2^a [#permalink]
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19 Nov 2014, 10:24
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Bunuel wrote: Tough and Tricky questions: Exponents. 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) \(a \lt c\) Kudos for a correct solution.\(m = 2^a3^b\) , \(n = 2^c\) \(p = \frac{2m}{n}\), now, substitute the value of m and n in the given expression. =\(\frac{2^{a+1}3^b}{2^C}\) \(p=2^{a+1c}3^{b}\) now, p will be odd, if the power of 2 becomes zero. i.e. a+1c=0 or ca=1 1) st.1 \(a \lt b\). not sufficient, as no information is provided about c st.2 \(a \lt c\) \(= ca\gt0\) now as per the question, p is an integer. therefore, difference between c and a cannot be more than 1. because, if we consider that a=2 and c=4. then a+1c= 2+14=1 thus expression \(p=2^{a+1c} 3^{b}\), will become \(p=\frac{3^b}{2}\) which results in p not becoming an integer. hence difference cannot be more than 1. thus the only possible value of ca is 1. now recall from 1) if ca=1. then p is odd. Also value of b doesn't matter here because 3^b will always be odd. hence statement 2 alone is sufficient.



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Re: Positive integers a, b, c, m, n, and p are defined as follows: m = 2^a [#permalink]
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19 Nov 2014, 21:59
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P = 2m/n = 2*2^a*3^b/2^c = 2^(a+1c)*3^c [Note that a, b, c, m, n, and p are all +ve integers]
Therefore, to find out whether the number is odd or even we have to know the value of a+1c
Statement 1: a<b. Clearly insufficient since the we don't still know the value of a+1c
Statement 2: a < c. Therefore either a+1< c or a+1 = c
if a+1< c, then p can't be an integer Therefore, for p to be an integer a+1=c, in which case p = 3^c (which is always odd)
Hence sufficient
B) should be the answer



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Re: Positive integers a, b, c, m, n, and p are defined as follows: m = 2^a [#permalink]
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20 Nov 2014, 03:02
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Given that , m=(2^a)(3^b) ; n=2^c & p=2m/n
Lets find m/n=(2^a)(3^b)/2^c=(2^(ac))(3^b) so p=2m/n=2*(2^(ac))(3^b)................A
1)a<b; looking at equation A one can say that this statement does not solve the riddle,since there in no nexus between a & b found. INSUFFICIENT
2)a<c; As a & c are directly linked in the equation A,we can say from this statement that if a<c then ac will be a negative term & the whole ratio is multiplied by 2 so p=2*(2^(ac))(3^b) =even. SUFFICIENT
Answer B



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Re: Positive integers a, b, c, m, n, and p are defined as follows: m = 2^a [#permalink]
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20 Nov 2014, 07:41
is p=2m/n odd? =(2*2^a*3^b)/2^c =2^(a+1c)*3^b so is 2^(a+1c)*3^b odd?
statement 1 : a<b take a=1 and b=2 =2^(1+1c)*3^2 =2^(2c)*9
no information on c. If c=2, then 2^(22)*9 =2^0*9 =1*9 =9 odd
If c=1, then 2^(21)*9 =2^1*9 =2*9=18 even
statement 2 : a<c take a=1 and c=2 =2^(1+12)*3^b =2^0*odd (since b is positive integer, 3^b will always be odd integer) =1*odd =odd
take a=1 and c=3 =2^(1+13)*3^b =2^1*3^b =1/2*3^b not an integer
1+2 combined is also insufficient. Ans=E
OA please



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Re: Positive integers a, b, c, m, n, and p are defined as follows: m = 2^a [#permalink]
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20 Nov 2014, 07:52
Bunuel wrote: Tough and Tricky questions: Exponents. 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) \(a \lt c\) Kudos for a correct solution. 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.
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Re: Positive integers a, b, c, m, n, and p are defined as follows: m = 2^a [#permalink]
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21 Aug 2016, 03:57
Nice Question here P=2^ab+1 *3^b so in order for it to be even => ac+1≥1 => ac≥0 now statement 1 is insufficient as there is no clue of c statement 2 says that ac<0 hence the answer to the question is NO P CAN NEVER BE EVEN Sufficient Smash that B
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Re: Positive integers a, b, c, m, n, and p are defined as follows: m = 2^a [#permalink]
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02 May 2017, 07:22
B  p = 2 ^(a  c + 1) * 3 ^(b)
for p to be not odd . we need a  c + 1 < 1 (power of 2 to be negative) or b < 1 in A  we can't get anything b has nothing to do. more over even through B  Yes a < c => a  c < 0 => a  c +1 < 1 ; bingo, what we needed. Through this p is some factor and can never be some integer or odd number.
SO B.
Kudos if you like the explanation




Re: Positive integers a, b, c, m, n, and p are defined as follows: m = 2^a
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