Primes & Factors: Level 3

Solution (E): The sum is 40 (even). To get an even sum from three primes, one of them must be 2. So, \(2 + p + q = 40 \implies p + q = 38\). To maximize the product \(2 \times p \times q\), we need to maximize \(p \times q\). This happens when \(p\) and \(q\) are close to \(38/2 = 19\). The primes closest to 19 are 17 and 19. \(17+19=36\). (Close, but not 38). Let’s check other pairs: \(7 + 31 = 38\). Product = \(2 \times 7 \times 31 = 434\). Next pair: \(11 + 27\) (27 not prime). Next pair: \(5 + 33\) (33 not prime). Next pair: \(17 + 21\) (21 not prime). Wait, I missed 17 and 19. \(17+19 = 36\). I need 38. Let’s re-check \(p+q=38\). Primes are {7, 31}, {5, 33} (no), {3, 35} (no), {11, 27} (no), {17, 21} (no). Let’s check my {17, 19} logic. \(17+19=36\). Let’s re-check the primes {2, 7, 31}. Sum=40. Product=434. Let’s re-check primes {2, 17, 21} (no). {2, 11, 27} (no). Ah, {2, 19, 19}. But they must be *distinct*. So the only pair for 38 is {7, 31}. Wait, did I miss one? \(p+q=38\). Primes: 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37. \(7 + 31 = 38\). Product = \(2 \times 7 \times 31 = 434\). This seems to be the only one. *Rethink:* My Q1 explanation for E=646 used {2, 17, 19}. But \(2+17+19 = 38\). The question says the sum is 40. So the primes must be {2, 7, 31}. Product = 434. The answer is C.

Solution (C): The sum is 40 (even). For the sum of three primes to be even, one of them *must* be 2 (the only even prime). So, \(2 + p + q = 40 \implies p + q = 38\). We need to find two distinct primes \(p, q\) that sum to 38. We test pairs: \(p=3, q=35\) (Not prime) \(p=5, q=33\) (Not prime) \(p=7, q=31\) (Both prime!) \(p=11, q=27\) (Not prime) \(p=13, q=25\) (Not prime) \(p=17, q=21\) (Not prime) \(p=19, q=19\) (Not distinct) The only set of primes is {2, 7, 31}. The product is \(2 \times 7 \times 31 = 14 \times 31 = 434\).

Solution (A): Any prime number \(p > 3\) can be written as \(6k \pm 1\). In both cases, \(p^2 = (6k \pm 1)^2 = 36k^2 \pm 12k + 1 = 12k(3k \pm 1) + 1\). The term \(k(3k \pm 1)\) is always even, so \(12k(3k \pm 1)\) is a multiple of 24. Therefore, \(p^2 = 24m + 1\) for any prime \(p > 3\). So, \(p^2 – q^2 = (24m + 1) – (24j + 1) = 24m – 24j = 24(m-j)\). The result is always a multiple of 24, so the remainder is 0.

Solution (C): Factor out the smallest power, \(7^{18}\). \(n = 7^{18} (7^2 – 1)\) \(n = 7^{18} (49 – 1)\) \(n = 7^{18} (48)\) The prime factorization of 48 is \(3 \times 16 = 3 \times 2^4\). So, \(n = 7^{18} \times 3 \times 2^4\). The distinct prime factors are 2, 3, and 7. The greatest is 7.

Solution (E): 1. The prime factorization of 60 is \(2^2 \times 3^1 \times 5^1\). 2. For \((n+1)^2\) to be divisible by \(60\), it must contain all these prime factors. 3. \((n+1)^2\) must be divisible by \(3^1\). This means \(n+1\) must be divisible by 3. 4. \((n+1)^2\) must be divisible by \(5^1\). This means \(n+1\) must be divisible by 5. 5. \((n+1)^2\) must be divisible by \(2^2\). This means \(n+1\) must be divisible by 2. 6. Therefore, \(n+1\) must be divisible by 2, 3, and 5. 7. The least common multiple of {2, 3, 5} is \(2 \times 3 \times 5 = 30\).

Solution (A): Factor the difference of squares: \((x-y)(x+y) = 45\). The factor pairs of 45 are (1, 45), (3, 15), and (5, 9). Case 1: \(x-y=1\) and \(x+y=45\). Add the equations: \(2x = 46 \implies x=23\). Then \(y=22\). 22 is not prime. Case 2: \(x-y=3\) and \(x+y=15\). Add: \(2x = 18 \implies x=9\). 9 is not prime. Case 3: \(x-y=5\) and \(x+y=9\). Add: \(2x = 14 \implies x=7\). Then \(y=2\). Both 7 and 2 are prime, so this is the correct solution. \(x=7\).

Solution (D): 1. List the primes between 30 and 50: {31, 37, 41, 43, 47}. 2. Sum the numbers: \(31 + 37 + 41 + 43 + 47\). 3. \( (31+47) + (37+43) + 41 = 78 + 80 + 41 = 158 + 41 = 199 \). 4. We must find the prime factors of 199. \(\sqrt{199} \approx 14.1\). 5. Check primes up to 13: 199 is not divisible by 2, 3 (sum=19), 5. \(199 / 7 = 28 R 3\). \(199 / 11 = 18 R 1\). \(199 / 13 = 15 R 4\). Since 199 is not divisible by any prime \(\le 14\), 199 is a prime number. The largest (and only) prime factor of 199 is 199.

Solution (C): A) \(q+r\): e.g., \(2 + 3 = 5\) (prime). B) \(q-r\): e.g., \(5 – 2 = 3\) (prime). C) \(qr\): This is the product of two prime numbers. Its factors are \(1, q, r, qr\). By definition, a prime number has exactly two factors. A product of two primes will always have at least three factors (e.g., \(2 \times 3 = 6\). Factors: 1, 2, 3, 6). Therefore, \(qr\) cannot be prime. D) \(qr+1\): e.g., \(2 \times 3 + 1 = 7\) (prime). E) \(q+2\): e.g., \(3 + 2 = 5\) (prime).

Solution (B): Let \(P(k)\) be the set of distinct prime factors of \(k\). \(P(12) = \{2, 3\}\). \(P(15) = \{3, 5\}\). \(P(20) = \{2, 5\}\). We are given \(|P(12) \cup P(n)| = 3 \implies |\{2, 3\} \cup P(n)| = 3\). We are given \(|P(15) \cup P(n)| = 3 \implies |\{3, 5\} \cup P(n)| = 3\). To minimize \(|P(20) \cup P(n)|\), we should try to make \(P(n)\) small. Let \(P(n) = \{2, 5\}\). Check 1: \(|\{2, 3\} \cup \{2, 5\}| = |\{2, 3, 5\}| = 3\). (Correct) Check 2: \(|\{3, 5\} \cup \{2, 5\}| = |\{2, 3, 5\}| = 3\). (Correct) This is a valid \(P(n)\). Now find the count for \(20n\): \(|P(20) \cup P(n)| = |\{2, 5\} \cup \{2, 5\}| = |\{2, 5\}| = 2\).

Solution (A): To maximize \(x\), we must maximize \(a\) and minimize \(b\). The smallest prime \(b\) is 2. If \(b=2\), then \(x = a – 2\). Since \(x\) must be prime, \(x\) must be odd (as \(x > 2\)). \(a = x + 2\). We need to find the largest prime \(x\) such that \(x+2\) is also a prime (and \(< 40\)). The largest prime \(a < 40\) is 37. If \(a=37\), \(x = 37 - 2 = 35\). 35 is not prime. The next largest prime \(a\) is 31. If \(a=31\), \(x = 31 - 2 = 29\). 29 is prime. This gives a value of \(x=29\).

Solution (B): For \(n\) to be divisible by 6, 10, and 15, \(n\) must be a common multiple. We find the Least Common Multiple (LCM) of {6, 10, 15}. Prime factors: 6 = \(2 \times 3\) 10 = \(2 \times 5\) 15 = \(3 \times 5\) The LCM must have the highest power of each prime: \(2^1 \times 3^1 \times 5^1 = 30\). So, \(n\) must be a multiple of 30. The multiples < 200 are {30, 60, 90, 120, 150, 180}. The prime factors of 30 are {2, 3, 5}. The prime factors of any multiple of 30 (like 60, 90, etc.) must *at least* include {2, 3, 5}. This means \(n\) must have *at least* 3 prime factors. The question is "how many prime factors does n have?". This is ambiguous. It could be 3 (if n=30) or 4 (if n=210, but n<200) or 4 (if n=30*7=210). Let's re-read the image Q5. "how many different positive prime factors?". So, how many does \(n=30, 60, 90, 120, 150, 180\) have? PF(30) = {2, 3, 5} -> 3 PF(60 = 30*2) = {2, 3, 5} -> 3 PF(90 = 30*3) = {2, 3, 5} -> 3 PF(120 = 30*4) = {2, 3, 5} -> 3 PF(150 = 30*5) = {2, 3, 5} -> 3 PF(180 = 30*6) = {2, 3, 5} -> 3 In all possible cases, \(n\) has exactly 3 distinct prime factors.