GMAT Primes & Factors: Level 1

Solution (C): The first five prime numbers are 2, 3, 5, 7, and 11. Their sum is \(2 + 3 + 5 + 7 + 11 = 28\).

Solution (D): The number 2 is the only even prime number. Every other prime number (3, 5, 7, 11, etc.) must be odd.

Solution (A): The prime factorization of 30 is \(2 \times 3 \times 5\). The distinct prime factors are 2, 3, and 5.

Solution (B): A prime number greater than 2 must be odd. The expression \(p+q\) is therefore \(\text{Odd} + \text{Odd}\). The sum of two odd numbers is always even.

Solution (D): 9 is divisible by 3. 15 is divisible by 3 and 5. 21 is divisible by 3 and 7. 27 is divisible by 3 and 9. 23 is only divisible by 1 and 23, so it is prime.

Solution (A): The prime factorization of 100 is \(10 \times 10 = (2 \times 5) \times (2 \times 5) = 2^2 \times 5^2\). The distinct prime factors are 2 and 5. The largest is 5.

Solution (C): \(p=2\) and \(q\) is an odd prime. A) \(p+q = 2 + \text{Odd} = \text{Odd}\). If \(q=3\), \(2+3=5\) (prime). If \(q=5\), \(2+5=7\) (prime). If \(q=7\), \(2+7=9\) (not prime). B) \(pq = 2q\). This is always even and greater than 2, so it can never be prime. C) \(q-p = \text{Odd} – 2 = \text{Odd}\). If \(q=5\), \(5-2=3\) (prime). If \(q=7\), \(7-2=5\) (prime). If \(q=13\), \(13-2=11\) (prime). This is possible. *Correction:* Both A and C are possible. Let’s re-read the Q2 from the image. “CANNOT be”. Let’s change my Q7 to “which of the following CANNOT be a prime number?”. In that case, B (\(pq\)) and D (\(2q\)) are always even. Let’s stick to “could be” and fix the options. A) \(p+q\). Possible. B) \(pq\). Impossible (2q is even). C) \(q-p\). Possible. D) \(2q\). Impossible (even). E) \(q+p+1 = 2+q+1 = q+3\). \(q=\text{Odd}\), so \(q+3 = \text{Odd}+\text{Odd} = \text{Even}\). Impossible. This leaves A and C. Let’s change A to \(p+q+1\). \(\text{Odd}+2+1 = \text{Odd}+3 = \text{Even}\).

Solution (C): \(p=2\) and \(q\) is an odd prime. A) \(p+q+1 = 2 + \text{Odd} + 1 = 3 + \text{Odd} = \text{Even}\). Cannot be prime. B) \(pq = 2q\). This is always even and greater than 2. Cannot be prime. C) \(q-p = \text{Odd} – 2 = \text{Odd}\). This *could* be prime. (e.g., if \(q=5\), \(5-2=3\), which is prime). D) \(2q\). Cannot be prime. E) \(q+3 = \text{Odd} + \text{Odd} = \text{Even}\). Cannot be prime.

Solution (B): 1. Find the prime factorization of 12: \(12 = 4 \times 3 = 2^2 \times 3^1\). 2. We have \((2^2 \times 3^1) \times n = \text{Square}\). 3. For a perfect square, all exponents must be even. 4. \(2^2\) is fine. \(3^1\) is odd. We need to multiply by \(3^1\) to get \(3^2\). 5. The smallest value for \(n\) is 3.

Solution (D): The two smallest prime numbers are 2 and 3. \(a = 2\) \(b = 3\) \(b – a = 3 – 2 = 1\).

Solution (A): We check the integers after 20. 21 is divisible by 3. 22 is even. 23 is not divisible by 2, 3, 5… It is prime.