Can any smart anons help me out with the first 4 question?

Question:Can any smart anons help me out with the first 4 question?

Answer: Nvm, 13 is a divisor of 39. I should be fired

Answer: Hahah thanks for trying. Do you reckon you could at least try 1d? Well drink to the answer

Answer: Do you mean 1a to 1d or 1 to 4?

Answer: (1a) Modulo 3 you obtain x, x+1, x+2, so one of them must be divisible by 3. The only prime with this property is 3 which implies x = 3 but then x+32 = 35 is not prime. A contradiction. (1c) consider n modulo 12. If n odd and not divisible by 3 then you have n congruent to ±1 or ±5 (mod 12). n = ±1 (mod 12) = n&2 = 1 (mod 12) n = ±5 (mod 12) = n&2 = 25 = 1 (mod 12) (1d) is true, consider n! + i, 2≤ i ≤ n. n!+2 is divisible by 2, n!+3 is divisible by 3, and so on, n!+n is divisible by n. This means that there are arbitrarily long sequences of consecutive composite numbers, and with say n = 101 you obtain the special case of 100 consecutive composite numbers.

Answer: I think (1b) is true if c is relatively prime to m but not in general. Consider for example a = 1, b = 5, c = 2, m = 4. a = b = 1 (mod 4) 2&1 = 2 (mod 4) 2&5 = 32 = 0 (mod 4)

Answer: Question 1 part d

Answer: Thank you so much anon! May comfy sleeps grace you.

Answer: Wait so is 1c true or false?

Answer: To any other anons, help on the rest would be welcome.

Answer: All good anons I got that one

Answer: If any anons still here, just stuck on question 2 Assume that there is some prime number p10 for which 7 does not divide (p&2+3)(p&2+5)(p&2+6). Now, (p&2+3)(p&2+5)(p&2+6) expands to p&6+14p&4+14p&2+90. When you take modulo 7, this simplifies to p&6+6. So, if 7 does not divide (p&2+3)(p&2+5)(p&2+6), this implies that p&6 (mod 7) is not congruent to 1. However, n&6 (mod 7) =1 unless n (mod 7) = 0. Since p is a prime number greater than 10, p (mod 7) != 0, therefore 7 must divide (p&2+3)(p&2+5)(p&2+6). There's a general rule that for any prime number p, n&(p-1) is congruent to 1 modulo p for all n not divisible by p. I'm not sure if you're at the point where you can just use that rule, or if you'd have to prove it from scratch. In any case, it's easily demonstrated for 7 just through brute force.

Answer: you can solve this also as follows. Let p 10 be prime and assume that 7 does not divide (p&2+3)(p&2+5)(p&2+6). Consider p mod 7: p mod 7 = 0 is not possible since p is a prime 10. p mod 7 = ± 1 = p&2 + 6 mod 7 = 0 p mod 7 = ± 2 = p&2 + 3 mod 7 = 0 p mod 7 = ± 3 = p&2 + 5 mod 7 = 0 Therefore, the term (p&2+3)(p&2+5)(p&2+6) is divisible by 7, a contradiction.

Answer: Actually, we don't need the assumptions p10 prime, we just need that p is not divisible by 7. For example, if p = 1 then (4)(6)(7) is divisible by 7. If p = 4 then (19)(21)(22) is divisible by 7, and so on. Or we can reduce the term to just p&2 + 6, as used in 709295 .

Answer: *p&6 + 6