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I believe the answer should be C, S1 tells us that X is an even number greater than 3 so we have 4,12,28. (As \sqrt{(x-3)} is also an odd number So not sufficient S2 tells us that X is a multiple of 3. so we have 3,6,9... so not sufficient.

S1 and S2 tells us that X is an even multiple of 3 that is 12,36,

However, S1 says \sqrt{x-3} is odd. It does not however say that it is an integer or not.

for eg. the number 18 will satisfy both the both S1 and S2. As Sq root of 15 is odd and 18 is a multiple of 3. In fact sq of any odd number will be odd. Unless we assume that the statement "sq root of a number is odd" means that the number is a perfect sq. Can someone please clarify if that this is a valid assumption that we can make for all gmat questions?

However, S1 says \sqrt{x-3} is odd. It does not however say that it is an integer or not.

for eg. the number 18 will satisfy both the both S1 and S2. As Sq root of 15 is odd and 18 is a multiple of 3. In fact sq of any odd number will be odd. Unless we assume that the statement "sq root of a number is odd" means that the number is a perfect sq. Can someone please clarify if that this is a valid assumption that we can make for all gmat questions?

Only an integer can be odd. i.e. if sqrt{x-3} is odd then it must be a number. So 18 do not satisfies.

However, S1 says \sqrt{x-3} is odd. It does not however say that it is an integer or not.

for eg. the number 18 will satisfy both the both S1 and S2. As Sq root of 15 is odd and 18 is a multiple of 3. In fact sq of any odd number will be odd. Unless we assume that the statement "sq root of a number is odd" means that the number is a perfect sq. Can someone please clarify if that this is a valid assumption that we can make for all gmat questions?

Only an integer can be odd. i.e. if sqrt{x-3} is odd then it must be a number. So 18 do not satisfies.

Let sqrt(x-3) =n with n being odd. <=> x-3 = n^2 <=> x = n^2 + 3

Going through odd integers we get x = 4, 12, 28, 52, 84, 124, 172 etc.

=> INSUFF.

Statement 2: x is a multiple of 3 => INSUFF.

Both statements combined:

Looking at the list of possible values for x above, we see that all values that are divisible by 3 are divisible by 4, too. Therefore, these values are also a multiple of 12.

(1) \(\sqrt[]{x−3}\) is odd => \(\sqrt[]{x−3}\) = a => a= odd integer => \(x-3 = a^2\) => \(x= a^2+3\)

=> here x is an integer For a as an odd integer, for a=1,3,5,7,9.... value of x = 4,12,28,52,84....... Insufficient

(2) x is a multiple of 3 => x is an integer and x =3b => here b is an integer => x=3,6,9,12,15,18,21,24...... Clearly Insufficient

1+2 \(x= a^2+3\) and x =3b , here a is an odd integer and b is an integer => \(3b=a^2+3\) => b= \(\frac{a^2}{3}\)+1 Here as 'b' need to be an integer 'a' should be multiple of 3

=> a is odd and multiple pf 3 => \(x= a^2+3\) , a=3,9,15 => x=12,84,228...... => x is multiple of 12

As x is a multiple of 3, we need to see whether x is a also a multiple of 4. Let's say \(\sqrt{x-3} = k\) \(x = k^2 + 3 = (k - 1)(k+1) + 4\)

Because k is an odd number(we have that condition from \(\sqrt{x-3}\) is an odd number), that means \((k - 1)(k+1) + 4\) is multiple of 4. So we have x is multiple of 3 and 4, so x is multiple of 12