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

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

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

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.

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

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

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

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.

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

2(2a3b)2c --> 2a+1−c3b

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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a, b, c, m, n, and p are positive integers.

Since p = (2)(2^a)(3^b)/(2^c) = [2^(a+1)](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 [There are no integers between a and a + 1, which are two consecutive integers.]

Therefore, the answer to the rephrased question is a definite Yes.

Statement two is sufficient.

Answer: B