AAPL wrote: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 < b.
(2) a < c.
The OA is B.
Please, can any expert assist me with this DS question? I don't have it clear and I appreciate if any explain it for me. Thanks.
We have p = (2*2^a*3^b) / 2^c = 2^(1+a-c)*3^b
Since b is a positive integer, 3^b is odd irrespective of whether b is even or odd. So whether p is odd depends on the value of 2^(1+a-c).
Since p is an integer, the value of 2^(1+a-c) cannot be less than 1 or fraction. Thus, 1 + a ≥ c.
Case 1: If 1 + a = c, then 2^(1+a-c) = 2^0 = 1. Thus, 2^(1+a-c)*3^b = 1*3^b = 3^b = Odd
Case 2: If 1 + a > c, then 2^(1+a-c) = 2^(a positive integer) = even. Thus, 2^(1-a+c)*3^b = Even*3^b = Even
So, it depends on whether 1 + a = c or 1 + a > c.
(1) a < b
There is no information about c. Insufficient.
(2) a < c
=> 1 + a > c is not possible, thus, 1 + a = c. This imples that p is odd. Sufficient.
The correct answer:
B
Hope this helps!
-Jay
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