If x is a positive integer, is x prime?

said A for this one.

y^2 = 9,16,25,36 and so on.

All these numbers have more than two factors, and so x cannot be prime.

B says z=3,4,5,6 and so on. Some of the numbers are prime, and some are not, so this statement by itself is insufficient.

It says distinct factors, not distinct prime factors.

So, for example, distinct factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36: 9 factors.
Distinct factors of 100 are 1, 2, 4, 5, 10, 20, 25, 50, and 100: 9 factors.
Distinct factors of 4 are 1, 2, and 4: 3 factors.

Does this make sense?

Hi there,

Can you please explain the statement "The number of distinct factors of a positive perfect square is ALWAYS ODD"? e.g. distinct factors of 36 are 2 and 3 (=even). Likewise for 100 are 2 and 5 ( = even).

Thank you,

TO

1) SUFF.: If x has the same number of factors as y2, then x cannot be prime. A prime number is a number that has only itself and 1 as factors. But a square has at least 3 prime factors. For example, if y is prime, y = 2, then y2 = 4, which has 1, 2, and 4 as factors. If the root (in this case y) is not prime, then the square will have more than 3 factors. For example, if y = 4, then y2 = 16, which has 1, 2, 4, 8, and 16 as factors. In either case, x will have at least 3 factors, establishing it as nonprime.

(2) INSUFF.: If z is prime, then x will have only two factors, making it prime. But if z is nonprime, it will have either one (if z = 1) or more than two factors, which means x will have either one or more than two factors, making x nonprime. Since we do not know which case we have, we cannot tell whether x is prime

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