Question 476120
1. The sum of 11 consecutive integers x, x+1, ..., x+10 is 11x + 55, divisible by 11.


2. The sum of 12 consecutive integers x, x+1, ..., x+11 is 12x + 66. 12x ≡ 0 (mod 12), 66 ≡ 6 (mod 12) so 12x + 66 cannot be a multiple of 12.


3. Either n or 2n + 1 will be a multiple of 2 (if n is even or n is odd). Also, if n ≡ 0 (mod 3), then n is divisible by 3; if n ≡ 1 (mod 3), then 2n + 1 ≡ 0; if n ≡ 2 (mod 3) then 7n + 1 ≡ 0. Hence, for all n, n(2n+1)(7n+1) is divisible by 6.


4. We know that n(2n+1)(7n+1) is divisible by 3 for all n from the previous problem, so we can check this expression modulo 4. If n ≡ 0 (mod 4), we are done; if n ≡ 1 (mod 4) then 7n+1 ≡ 0, and n ≡ 2 or n ≡ 3 result in the expression not being divisible by 4. All n that are congruent to 0 or 1 mod 4 will result in the expression being divisible by 12.