nades09 wrote:

Thanks!

(1) The only squares that have 3 distinct positive factors are 4, 25 and 49.

The factors are 1,x,\sqrt{x}.

Since \sqrt{x} for the above three numbers needs to be considered, hence there will be three different values for the sums

Hence (1) is insufficient

(2) x^2-1 = 3k where k =odd integer

Hence, x^2 =3k+1 is even

There can be multiple values where x^2 is even

Hence (2) is insufficient

(1)+(2) - x^2 should be even and should have 3 distinct positive factors

The only value that satisfies both conditions is 4

\sqrt{x} is 2, hence we can find the sum

Ans: C

You got most of it. If you go through factors theory, it will help you understand that only a square of a prime number can have 3 factors. e.g. 4 or 25 or 49 or 121 or 169..

1 , 2, 4 are factors of 4

1, 5, 25 are factors of 25

1, 7, 49 are factors of 49

1, 11, 121 are factors of 121 etc

So if

x has 3 factors,

\sqrt{x} must be prime.

Also from statement 2,

x^2 = 3k + 1 where k is odd. So 3k is odd and 3k + 1 is even. So

x^2 is even. Now, if

x^2 is even, x has to be even too (It is not possible that a power of an odd number becomes even. If x is odd,

x^2, x^3 etc all will be odd. If x is even,

x^2, x^3etc all will be even.). Then

\sqrt{x} must also be even.

The only number that is even and prime is 2. So

\sqrt{x} must be 2.

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