VeritasPrepKarishma wrote:
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, \(x^2, x^3\)etc 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.
I thought statement 2 was sufficient as \(\sqrt{x}\) is supposed to be an integer as only integer can have factors .
Based on this, statement 2 says product of 2 consecutive odd integers is odd
(x-1)(x+1)= odd
1*3= 3 here x= 2 but \(\sqrt{2}\) is not an integer hence we cannot take x= 2
3*5 = 15 here x= 4 and \(\sqrt{4}\) is an integer hence we can take x= 4
5*7 = cannot take, as this is not in the form 3K
7*9=63 here x = 8 but \(\sqrt{8}\) is not an integer hence we cannot take x= 8
9*11=99 X=10 but \(\sqrt{10}\) is not an integer hence we cannot take x= 10
so from statement 2 after testing various numbers
I felt that x = 4 was the only value that qualified hence B was the answer.
Are there any values that I missed ?
or \(\sqrt{x}\) need not be an integer?
Here is another value: \(x^2 = 3*85 + 1 = 256\) --> \(x = 16\) --> \(\sqrt{x}=4\).