jainrahul1985 wrote:If the prime numbers p and t are the only prime factors of the integer m, is m a multiple of p^2t ?
(1) m has more than 9 positive factors.
(2) m is a multiple of p^3.
OA B
This is a yes/no DS question: the possible answers here are
Yes - m is a multiple of p^2t, or
No - m is not a multiple of p^2.
Your initial approach for these questions is usually one of elimination: try to show that the statements allow both a yes and a no.
Use numerical examples to reduce the problem to a manageable state: let's say that p and t are 2 and 3 (two primes).
(1) says that m has more than 9 factors. Start with a basic model of 2*3 = 6: 6 has only four factors: 1, 2, 3, 6. Evidently we need to add more "building blocks" to m to reach the 9 factor requirement. Since m has only 2 and 3 as prime factors, that's the only thing we CAN add - more powers of 2, or 3, or both.
So what it comes down to is this: who is 2 and who is 3? This is the "wriggle room" that allows you to find examples of both a yes and a no:
fix p as 2 and t as 3. You can keep just one power of 2 and add as many powers of 3 as needed to meet the 9 factors requirement (for example, 2*3^100 will have much more than 9 factors) - in which case m will NOT be a multiple of p^2, since there's only one power of 2, OR
you can do the opposite: keep the t=3 as a single power, and add as many powers of 2 as needed, making sure that m WILL be divisible by p^2t.
Since you have both a yes and a no answer, stat. (1) is insufficient.
(2) fixes the p^2 part: if m is divisible by p^3, it is definitely divisible by a smaller power of p^2. We also know from the question stem that m is divisible by t (since p and t are prime factors of m), so the combination of (2) and the stem means that m MUST be a multiple of p^2t, and the answer is a definite "yes". Sufficient.
