Primes & Factors: Level 2

Solution (D): The prime numbers between 20 and 40 are: 23, 29, 31, 37. Their sum is \(23 + 29 + 31 + 37 = 120\).

Solution (B): A composite number is a positive integer with more than two factors (1 and itself). The expression \(pq\) is the product of two prime numbers. Its factors are 1, \(p\), \(q\), and \(pq\). Since \(p\) and \(q\) are distinct, this number always has at least four factors, so it must be composite. (e.g., \(p+q\) could be \(2+3=5\), which is prime).

Solution (A): Factor out the smallest power of 3, which is \(3^{10}\). \(n = 3^{10}(1 + 3^2)\) \(n = 3^{10}(1 + 9)\) \(n = 3^{10}(10)\) \(n = 3^{10} \times (2 \times 5)\) The distinct prime factors are 2, 3, and 5.

Solution (D): 1. The prime factorization of 12 is \(2^2 \times 3\). 2. For \(\frac{n^{10}}{2^2 \times 3}\) to be an integer, \(n^{10}\) must be divisible by \(2^2\) and \(3\). 3. If \(n^{10}\) is divisible by 3, \(n\) must be divisible by 3. 4. If \(n^{10}\) is divisible by \(2^2\), \(n\) must be divisible by 2. 5. Therefore, \(n\) must have prime factors of 2 and 3. 6. If \(n\) is divisible by 2 and 3, it must be divisible by their least common multiple, which is 6.

Solution (B): Factor out the smallest power, \(5^{15}\). \(5^{15}(5^2 – 1)\) \(5^{15}(25 – 1)\) \(5^{15}(24)\) Find the prime factors of 24: \(24 = 8 \times 3 = 2^3 \times 3\). The full expression is \(5^{15} \times 2^3 \times 3\). The distinct prime factors are 2, 3, and 5. The greatest is 5.

Solution (E): To get the largest sum, we must add the two largest primes that are less than 50. The primes less than 50 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47. The two largest are 47 and 43. The sum is \(47 + 43 = 90\).

Solution (D): We must test a prime \(w > 3\). Let’s use \(w = 5\). A) \(w = 5\) (Prime) B) \(w-2 = 5-2 = 3\) (Prime) C) \(w+2 = 5+2 = 7\) (Prime) D) \(w+4 = 5+4 = 9\). 9 is composite. This is possible. E) \(w-4 = 5-4 = 1\) (Neither prime nor composite) Let’s test \(w=7\). D) \(w+4 = 7+4 = 11\) (Prime). Let’s test \(w=11\). D) \(w+4 = 11+4 = 15\) (Composite). Since it’s possible, D is the answer.

Solution (C): We need to find two primes \(a, b < 20\) that sum to a prime \(x\). Primes < 20 are {2, 3, 5, 7, 11, 13, 17, 19}. If we add two odd primes, the sum is even (e.g., \(19+17=36\)), which cannot be prime (since it's > 2). Therefore, one of the primes *must* be 2. Let \(a=2\). We need to find the largest prime \(b < 20\) such that \(2+b\) is also prime. Let \(b = 19\) (largest prime < 20). \(x = 2 + 19 = 21\). 21 is not prime. Let \(b = 17\). \(x = 2 + 17 = 19\). 19 is prime. This is a possible value for \(x\).

Solution (C): The single-digit prime numbers are {2, 3, 5, 7}. We check the options to see which number is formed using only these digits. (A) 21: 1 is not prime. (B) 34: 4 is not prime. (C) 57: 5 is prime and 7 is prime. This is a possible number. (D) 13: 1 is not prime. (E) 43: 4 is not prime.

Solution (A): We find the prime factorization of \(n\). \(n = (2 \times 5) \times (20) \times (30)\) \(n = (2 \times 5) \times (2^2 \times 5) \times (2 \times 3 \times 5)\) Combine the factors: \(n = 2^{(1+2+1)} \times 3^1 \times 5^{(1+1+1)} = 2^4 \times 3^1 \times 5^3\) The distinct (different) prime factors are 2, 3, and 5. There are 3 of them.