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 #326495 by tinbar(133) $\"\"$ $\"About$

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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\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "for the thid one/fourth, before I answer, I need to know whether you are familiar with and are allowed to use modular arithmetic \n" ); document.write( "

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