document.write( "Question 97923This question is from textbook Intro to Real Analysis
\n" ); document.write( ": Prove that n^3 + (n+1)^3 + (n+2)^3 is divisible by 9 for all n in Natural numbers. I have tried, let mathematical induction, let P(n) be the statement, then then n^9 - n is divisible by 9. I'm stuck...\r
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Algebra.Com's Answer #71231 by aaaaaaaa(138)\"\" \"About 
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\"x mod y\" is short for \"remainder of the division of x by y\". Iff \"x mod y\" is 0, x is divisible by y. In our case, y = 9.\r
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\n" ); document.write( "\n" ); document.write( "\"n%5E3+%2B+%28n%2B1%29%5E3+%2B+%28n%2B2%29%5E3\" (mod 9) =
\n" ); document.write( "\"n%5E3+%2B+n%5E3+%2B+3n%5E2+%2B+3n+%2B+1+%2B+n%5E3+%2B+6n%5E2+%2B+12n+%2B+8\" (mod 9)=
\n" ); document.write( "\"3n%5E3+%2B+9n%5E2+%2B+15n+%2B+9\" (mod 9) =
\n" ); document.write( "\"3n%5E3+%2B+9n%5E2+%2B+9n+%2B+6n+%2B+9\" (mod 9)\r
\n" ); document.write( "\n" ); document.write( "9n^2 is always divisible by 9, so we shouldn't worry about that term. The same goes for the \"+9\" at the end, and the \"+9n\" at the middle.\r
\n" ); document.write( "\n" ); document.write( "\"3n%5E3+%2B+cross%289n%5E2%29+%2B+cross%289n%29+%2B+6n+%2B+cross%289%29\" (mod 9) =
\n" ); document.write( "\"3n%5E3+%2B+6n\" (mod 9)\r
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\n" ); document.write( "\n" ); document.write( "Therefore,
\n" ); document.write( "P(x) = \"n%5E3+%2B+%28n%2B1%29%5E3+%2B+%28n%2B2%29%5E3\" (mod 9) = 0
\n" ); document.write( "is equivalent to
\n" ); document.write( "S(x) = \"3n%5E3+%2B+6n\" (mod 9) = 0\r
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\n" ); document.write( "\n" ); document.write( "Now suppose S(n) is true. I'll try to prove that S(n+1) is also true.\r
\n" ); document.write( "\n" ); document.write( "\"3%28n%2B1%29%5E3+%2B+6%28n%2B1%29\" (mod 9) =
\n" ); document.write( "\"3%28n%5E3+%2B+3n%5E2+%2B+3n+%2B+1%29+%2B+6n+%2B+15\" (mod 9) =
\n" ); document.write( "\"3n%5E3+%2B+cross%289n%5E2%29+%2B+cross%289n%29+%2B+3+%2B+6n+%2B+15\" (mod 9) =
\n" ); document.write( "\"3n%5E3+%2B+6n+%2B+cross%2818%29\" (mod 9) =
\n" ); document.write( "\"3n%5E3+%2B+6n\" (mod 9)\r
\n" ); document.write( "\n" ); document.write( "But wait, if S(n) is true, then\r
\n" ); document.write( "\n" ); document.write( "\"3n%5E3+%2B+6n\" (mod 9) = \"0\" (mod 9)\r
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\n" ); document.write( "\n" ); document.write( "The inductive step is done. Now we need to check the base case of natural numbers (which is 0)\r
\n" ); document.write( "\n" ); document.write( "\"3%280%5E3%29+%2B+3%2A0\" (mod 9) =
\n" ); document.write( "\"0\" (mod 9)\r
\n" ); document.write( "\n" ); document.write( "S(x) is therefore true for all naturals.\r
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