document.write( "Question 745014: if n > 2 ,then show that n^5-5n^3+4n is divisible by 120 \n" ); document.write( "
Algebra.Com's Answer #454744 by tommyt3rd(5050)\"\" \"About 
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Clearly we assume that n is an integer...\r
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\n" ); document.write( "\n" ); document.write( "First let's factor:\r
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\n" ); document.write( "\n" ); document.write( "next we rearrange them is succession:\r
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\n" ); document.write( "\n" ); document.write( "\"%0D%0A%28n-2%29%28n-1%29n%28n%2B1%29%28n%2B2%29%0D%0A\"\r
\n" ); document.write( "\n" ); document.write( "which shows that this is a product of 5 consecutive numbers starting with n-2\r
\n" ); document.write( "\n" ); document.write( "and
\n" ); document.write( "\"%0D%0A120=12%2A10=2%5E3%2A3%2A5=3%2A5%2A8%0D%0A\"\r
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\n" ); document.write( "\n" ); document.write( "Now notice that if n>2 the each product is\r
\n" ); document.write( "\n" ); document.write( "1*2*3*4*5
\n" ); document.write( "2*3*4*5*6
\n" ); document.write( "3*4*5*6*7\r
\n" ); document.write( "\n" ); document.write( "and so on\r
\n" ); document.write( "\n" ); document.write( "each of which is a multiple of 3 (every third digit), 5 (every fifth digit), and 8 (there will always be a multiple of 2 and a multiple of 4).\r
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