document.write( "Question 476120: Show that the sum of 11 consecutive integers is always divisible by 11.\r
\n" ); document.write( "\n" ); document.write( "Show that the sum of 12 consecutive integers is never divisible by 12.\r
\n" ); document.write( "\n" ); document.write( "Show that n(2n + 1)(7n + 1) is divisible by 6 for all integers n.\r
\n" ); document.write( "\n" ); document.write( "Find all integers n such that n(2n + 1)(7n + 1) is divisible by 12. \n" ); document.write( "

Algebra.Com's Answer #326518 by richard1234(7193) $\"\"$ $\"About$

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.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( "

\n" );