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

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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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New post 19 Nov 2014, 08:22
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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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New post 19 Nov 2014, 11:24
3
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.
=\(\frac{2^{a+1}3^b}{2^C}\)
\(p=2^{a+1-c}3^{b}\)

now, p will be odd, if the power of 2 becomes zero. i.e. a+1-c=0 or c-a=1 -------------------------------------------1)

st.1 \(a \lt b\). not sufficient, as no information is provided about c

st.2 \(a \lt c\)
\(= c-a\gt0\)

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+1-c= 2+1-4=-1
thus expression \(p=2^{a+1-c} 3^{b}\), 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 c-a is 1.
now recall from 1) if c-a=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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New post 19 Nov 2014, 22:59
1
P = 2m/n = 2*2^a*3^b/2^c = 2^(a+1-c)*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+1-c

Statement 1: a<b. Clearly insufficient since the we don't still know the value of a+1-c

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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New post 20 Nov 2014, 04:02
1
Given that , m=(2^a)(3^b) ; n=2^c & p=2m/n

Lets find m/n=(2^a)(3^b)/2^c=(2^(a-c))(3^b)
so p=2m/n=2*(2^(a-c))(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 a-c will be a negative term & the whole ratio is multiplied by 2
so p=2*(2^(a-c))(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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New post 20 Nov 2014, 08:41
is p=2m/n odd?
=(2*2^a*3^b)/2^c
=2^(a+1-c)*3^b
so is 2^(a+1-c)*3^b odd?

statement 1 : a<b
take a=1 and b=2
=2^(1+1-c)*3^2
=2^(2-c)*9

no information on c.
If c=2, then 2^(2-2)*9
=2^0*9
=1*9
=9
odd

If c=1, then 2^(2-1)*9
=2^1*9
=2*9=18
even

statement 2 : a<c
take a=1 and c=2
=2^(1+1-2)*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+1-3)*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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New post 20 Nov 2014, 08:52
1
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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New post 21 Aug 2016, 04:57
Nice Question here P=2^a-b+1 *3^b
so in order for it to be even => a-c+1≥1 => a-c≥0

now statement 1 is insufficient as there is no clue of c
statement 2 says that a-c<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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New post 02 May 2017, 08:22
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
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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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New post 12 Jan 2019, 01:27
Let's analyse the question statement.

p will be odd only in the following cases:
a) n=2 and m is odd
b) n=\(2^x\) and m=odd integer*\(2^(x-1)\). If say \(n=2^4\) and m=odd integer*\(2^2\), then in case of p, we would have a 2 remaining in denominator which cannot be the case since p is an integer.

Now, let's analyse statements.

1) \(a \lt b\). Since we don't know what is the value of c, it is possible that p is even or odd. Let's take an example.

\(m = 2^23^3\). If c=1, then p=\(\frac{(2*2^23^3)}{2^1}\). Then, p is even

\(m = 2^23^3\). If c=3, then p=\(\frac{(2*2^23^3)}{2^3}\). Then, p is odd

Hence, statement 1 is insufficient

2) \(a \lt c\). Here we certainly know that a<c and the difference will only be by 1 (as discussed in the integer case of p in case b)

Thus, this statement will always give us p as odd. Hence, this statement is sufficient

Thus, answer is 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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New post 13 Jan 2019, 16:26
First, we simplify the question stem.
m=(2^a)(3^b)
n=(2^c)

p=2m/n=(2^a)(3^b) divided by (2^c)
=(2^a-c)*(3^b)

S2 answers this directly, p will be a fraction of some sort. a<c implies that 2 is raised to a negative power i.e. a fraction.
S1 tells us nothing much and hence is insufficient.

The answer would be B.
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Re: Positive integers a, b, c, m, n, and p are defined as follows: m = 2^a   [#permalink] 13 Jan 2019, 16:26
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