![\"\"](\"/images/stars/stars-13.gif\")
You can put this solution on YOUR website!
. A student thinks that P(n) = n^3 - n is always a multiple of 6 for all natural numbers n. What do you think? Provide a mathematical argument to show that the student is correct or a counter example to show that the student is wrong.
\n" ); document.write( "LET.... M = N^3-N=N(N^2-1)=N(N-1)(N+1)=(N-1)N(N+1)
\n" ); document.write( "THAT IS PRODUCT OF 3 CONSECUTIVE NUMBERS.THERE ARE 2 POSSIBILITIES
\n" ); document.write( "1...N IS EVEN..THEN 2 IS A FACTOR OF M.
\n" ); document.write( "2.OR N IS ODD..THEN N-1 IS EVEN AND HENCE 2 IS A FACTOR OF M
\n" ); document.write( "NOW AGAIN W.R.T DIVISION BY 3,THREE POSSIBILITIES ARISE..
\n" ); document.write( "1. N-1 IS DIVISIBLE BY 3 AND HENCE LEAVES A REMAINDER OF 0 ON DIVISION BY 3.
\n" ); document.write( "HENCE 3 IS A FACTOR OF M.
\n" ); document.write( "1. N-1 IS NOT DIVISIBLE BY 3 AND HENCE LEAVES A REMAINDER OF 1 SAY ON DIVISION BY 3.THEN N-1+2 = N+1 IS DIVISIBLE BY 3.HENCE 3 IS A FACTOR OF M.
\n" ); document.write( "1. N-1 IS NOT DIVISIBLE BY 3 AND HENCE LEAVES A REMAINDER OF 2 SAY ON DIVISION BY 3.THEN N-1+1=N IS DIVISIBLE BY 3.HENCE 3 IS A FACTOR OF M.
\n" ); document.write( " THUS IN ALL CASES 2 AND 3 ARE FACTORS OF M .SO 2*3=6 IS A FACTOR OF M
\n" ); document.write( "OR M IS A MULTIPLE OF 6.
\n" ); document.write( " IF YOU ARE CONVERSANT WITH MODULAR ARITHMATIC , THIS CAN BE PROVED MORE ELEGANTLY.
\n" ); document.write( "AT N=1,WE GET M=0...
\n" ); document.write( "AT N=2,WE GET M=6..ETC.. \n" ); document.write( "