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Bunuel
If integer p is a factor of 42, is p a prime number?

(1) p > 6
(2) p < 21


DS21228

P is a factor of 42

Factors of 42 = {1, 2, 3, 6, 7, 12, 14, 21, 42}

Prime factors of 42 = {2, 3, 7}


Question: Is P = one of {2, 3, 7}?

Statement 1: p > 6

i.e. p could be 7 (YES) or 14 (NO) hence

NOT SUFFICIENT

Statement 2: p < 21

i.e. p could be 7 (YES) or 14 (NO) hence

NOT SUFFICIENT

Combining the statements

6 < p < 21

Statement 1: p > 6

i.e. p could be 7 (YES) or 14 (NO) hence

NOT SUFFICIENT

Answer: Option E
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I marked this question wrong but thanks all for the explanation.

I wrote 42 as 2*3*7 (in multiples of prime number)
I missed the trick that factors could be a combination of prime factors as well (7, 14, 21) and considered only 7 and marked A as correct choice.

But I did consider 7, 6 (3*2), 14 (2*7) while rejecting B (p<21)

Silly mistake :(
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Bunuel
If integer p is a factor of 42, is p a prime number?

(1) p > 6
(2) p < 21


DS21228

Factor pairs:
\(1, 42\)
\(2, 21\)
\(6, 7\)
\(14, 3\)

(1) If \(p > 6\), then p can be \(7, 14, 21, 42\). p can be non-prime or prime. INSUFFICIENT.

(2) If \(p < 21\), p can be \(1, 2, 3, 6, 7, 14\). p can be non-prime or prime. INSUFFICIENT.

(1&2) Combined, p can be 7 or 14. INSUFFICIENT.

Answer is E.
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