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

Question

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 .

manpreetsingh86 · Accepted Answer

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.

kinghyts · Answer

P = 2m/n = 2*2^a*3^b/2^c = 2^(a+1-c)*3^c

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

akumar5 · Answer

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

desaichinmay22 · Answer

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

Bunuel · Answer

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.

stonecold · Answer

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

Nightmare007 · Answer

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

GMATin · Answer

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

pineapple123456 · Answer

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.

flawlessgmat · Answer

can someone explain the basic concept here. what are the rules for this to convert this to that?

2(2a3b)2c --> 2a+1−c3b

Bunuel · Answer

Operations involving the same bases:

Keep the base, add or subtract the exponent (add for multiplication, subtract for division)

a^n*a^m=a^{n+m}

\frac{a^n}{a^m}=a^{n-m}

Hence,  \frac{2(2^a*3^b)}{2^c} = \frac{2^{a+1}*3^b}{2^c}=2^{a + 1 - c}3^b .

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JeffTargetTestPrep · Answer

a, b, c, m, n, and p are positive integers. Since p = (2)(2^a)(3^b)/(2^c) = (3^b)/(2^c) = integer, it must be true that: c ≤ a + 1 We need to answer the question: Is p odd? Using that p is odd if and only if the 2s cancel out, we can rephrase the question as: Is c = a + 1 ? Statement One Alone: => a < b The relation between a and b doesn’t have any impact on the relation between a and c, which is our focus. Statement one is not sufficient. Eliminate answer choices A and D. Statement Two Alone: => a < c If we combine statement two with the initial constraint that c ≤ a + 1, we have: a < c ≤ a + 1 Since a and c are integers, it must be true that: c = a + 1 Therefore, the answer to the rephrased question is a definite Yes. Statement two is sufficient. Answer: B

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

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