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)![]() ![]() You can put this solution on YOUR website! Clearly we assume that n is an integer...\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "First let's factor:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( "\n" ); document.write( "next we rearrange them is succession:\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \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( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \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 \n" ); document.write( "\n" ); document.write( "Therefore they will always divide 120. \n" ); document.write( " \n" ); document.write( " |