document.write( "Question 1081675: Prove that $\"+n%5E3+%2B+5n+\"$ is divisible by 6, where n is any positive integer. \n" ); document.write( "

Algebra.Com's Answer #695728 by ikleyn(52838) $\"\"$ $\"About$

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\n" ); document.write( "Prove that $\"+n%5E3+%2B+5n+\"$ is divisible by 6, where n is any positive integer.
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\n" ); document.write( "\n" ); document.write( "My proof consists of two parts.\r
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\n" ); document.write( "\n" ); document.write( "Part 1. For any integer n, $\"+n%5E3+%2B+5n+\"$ is divisible by 2.\r
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document.write( "    Consider consequently the case of even \"n\" and the case of odd \"n\".\r\n" );
document.write( "    In both cases, the statement is true, which is OBVIOUS.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Part 2. For any integer n, $\"+n%5E3+%2B+5n+\"$ is divisible by 3.\r
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document.write( "     =  +  =  + .\r\n" );
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document.write( "     is always multiple of 3, since it is the product of three consecutive integers.\r\n" );
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document.write( "     is always multiple of 3, by the obvious reason.\r\n" );
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document.write( "    So the statement of the Part 2 is proven, too.\r\n" );
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\n" ); document.write( "\n" ); document.write( "Thus the general statement follows regarding divisibility by 6.\r
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