Question 476120: Show that the sum of 11 consecutive integers is always divisible by 11.
Show that the sum of 12 consecutive integers is never divisible by 12.
Show that n(2n + 1)(7n + 1) is divisible by 6 for all integers n.
Find all integers n such that n(2n + 1)(7n + 1) is divisible by 12.
Found 2 solutions by tinbar, richard1234:
Answer by tinbar(133)   (Show Source):
You can put this solution on YOUR website! for the first one: let x be some number, then the 10 consecutive integers after x is given by (x+1),(x+2),...,(x+10). If we add up these terms we get 11x+55, which is obviously divisible by 11; 11x+55/11 = x+5
for the second one: follow something similar as the first example and see what goes wrong, you'll get a similar expression as 11x+55(infact, if you are smart, you will simply modify this term since you only have to add the next consecutive integer), and you'll see something about this expression will tell you it cannot be divided by 12 no matter what x we pick.
for the thid one/fourth, before I answer, I need to know whether you are familiar with and are allowed to use modular arithmetic
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! 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.
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