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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