Official Solution:


Notice that, \(n\) has exactly two positive factors simply means that \(n\) is a prime number, so its factors are 1 and \(n\) itself.

(1) \(\frac{n}{2}\) is one of the factors of \(n\). Since \(\frac{n}{2}\) cannot equal to \(n\), then \(\frac{n}{2}=1\), thus \(n=2\). Sufficient.

(2) The lowest common multiple of \(n\) and \(n + 10\) is an even number. If \(n\) is an odd prime, then \(n+10\) is also odd. The LCM of two odd numbers cannot be even, therefore \(n\) is an even prime, so 2. Sufficient.


Answer: D
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Dear Sir,

Kind explanation on statement 2 please?

Hi

To understand statement 2, one thing that must be clear is that if a number has exactly 2 positive factors, then it must be a prime number.
A prime number has 2 positive factors: 1 and n (i.e., 1 and itself) - and NO other factor.
Also you should know that lowest prime number is '2' which is the ONLY EVEN prime number. All other prime numbers:
3, 5, 7, 11, 13, .... are ODD.

Now Statement 2. Given that LCM of n and (n+10) is Even. Notice that n and (n+10) will either Both be Even or Both be Odd.
If both are odd, then their LCM will be Odd only (LCM of two Odd numbers can Never be Even).

This means both n and (n+10) have to be even, which is Only possible when n=2. So Sufficient.

Hope you got it.

Statement 2) If n=2 then we have 2 and 12, therefore their LCM= 12, but if we have n=3, then we have 3 and 13, therefore their LCM= 26, (both LCM`S are even) so now we know that (n) could be = 2 or 3. (Insufficient)

Is there something that I am doing wrong?

26 is not divisible by 3. The LCM of 3 and 13 is 39, not 26.
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Only prime no. have two factors.
Using stmn1: n would be 2
sufficient
stmnt2: LCM of 2 and 12 would be 2
D is correct.

Stem tells us that N = prime (only 2 factors)

Statement (1)
N/2 --> Only even prime =2 therefore n = 2 --> Sufficient


Statement (2)
Test values
N N+10
2 12 LCM = 12 Even
3 13 LCM = 39 Odd
5 15 LCM = 75 Odd

We can keep testing, but the constraints of number properties tell us that EvenInteger = (Odd or Even)*Even, and the only thing that satisfies this is 2.
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