document.write( "Question 1076171: R0: 3,6,9,12,15,18,21,...
\n" ); document.write( "R1: 1,4,7,10,13,16,19,...
\n" ); document.write( "R2: 2,5,8,11,14,17,20,...
\n" ); document.write( "a) Write down an expression for the value of the general term in each of the three prgressions. Show that the integer 1706836 lies in R1. I got R0:3n, R1:3n-2 and R2:3n-1. Then i did 3n-2=1706836 and showed that n=568946, so it lies in R1 and is the 568946th term in the series.
\n" ); document.write( "b) Use the Binomial Theorem to show that if x is a term in R1 or R2 then x^6 is in R1. This is the part i don't know how to work out.
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Algebra.Com's Answer #690841 by KMST(5328)\"\" \"About 
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a) I agree with your expressions for the the general term in each of the three sequences.
\n" ); document.write( "To show if any number belongs to one of them,
\n" ); document.write( "I would just add digits, using module 9 clock arithmetic. The number belongs in the sequence that contains the final sum.
\n" ); document.write( "If the ending sum is 0, 3, or 6, the number is divisible by 3, and belongs in R0.
\n" ); document.write( "If it is 1, 4, or 7, it belongs in R1; if 2,5, or 8, it belongs in R2.
\n" ); document.write( "For \"1706836\" ,
\n" ); document.write( "the final \"36\" part adds up to \"3%2B69\" , and that 9 is dropped.
\n" ); document.write( "The \"1\" and \"8\" add up to \"1%2B8=9\" , and that 9 is dropped.
\n" ); document.write( "Then we are left with \"7%2B0%2B6=13\" , and since \"1%2B3=4\" , we conclude that
\n" ); document.write( "when \"1706836\" is divided by \"9\" , the remainder is \"4\" .
\n" ); document.write( "So, \"1706836=9K%2B4=3%2A3K%2B6-2=3%283K%2B2%29-2\" for some integer \"K\" ,
\n" ); document.write( "and that shows that it belongs in R1.
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\n" ); document.write( "HOW AND WHY CLOCK ARITHMETIC MOD 9 WORKS:
\n" ); document.write( "The sum of the digits of a number divisible by 9 is 9 or a multiple of 9.
\n" ); document.write( "As you are adding,
\n" ); document.write( "you drop partial sums that are 9, and keep adding the rest of the digits.
\n" ); document.write( "Similarly, for any partial two-digit sum,
\n" ); document.write( "you replace the sum of the digits for the two-digit partial sum.
\n" ); document.write( "At the end, you should have a one-digit final sum from 0 to 8.
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\n" ); document.write( "b)The numbers in R1 and R2 are \"1\" away from a multiple of \"3\" .
\n" ); document.write( "We could write them all as \"3p+%2B-+1\" for some non-negative integer \"p\"
\n" ); document.write( ".
\n" ); document.write( "(If you - or your teacher - do not like that expression, you could use
\n" ); document.write( "\"3p+%2B+1\" for terms of R1, with \"p=n-1\" ,
\n" ); document.write( "and \"3n+-+1\" for terms of R2,
\n" ); document.write( "and write twice as much for your proof).
\n" ); document.write( "For \"x=3p+%2B-+1\" ,
\n" ); document.write( "\"x%5E6=%283p+%2B-+1%29%5E6=%283p%29%5E6+%2B-+6%2A%281%29%283p%29%5E5\"\"%22%2B%22\"\"15%283p%29%5E4%281%29%5E2+%2B-+20%283p%29%5E3%281%29%5E3\"\"%22%2B%22\"\"15%283p%29%5E2%281%29%5E4+%2B-+6%283p%29%281%29%5E5%2B1%5E6\"
\n" ); document.write( "\"%22=%22\"\"%283p%29%5E6+%2B-+6%283p%29%5E5\"\"%22%2B%22\"\"15%283p%29%5E4+%2B-+20%283p%29%5E3\"\"%22%2B%22\"\"15%283p%29%5E2+%2B-+6%283p%29+\" \"%22%2B+1%22\"
\n" ); document.write( "It should be obvious that all terms of that sum are multiples of \"3\" ,
\n" ); document.write( "except that \"%22+%2B+1%22\" at the end,
\n" ); document.write( "and that shows that \"x%5E6\" is a term in R1.
\n" ); document.write( "
\n" ); document.write( "If your teacher insisted on seeing the \"n\" that gives \"x%5E6\" the right to belong to R1,
\n" ); document.write( "you would have to write more, starting with getting out \"3\" as a common factor:
\n" ); document.write( "\"x%5E6+=+3\"\"%22%5B%22\"\"p%2A%283p%29%5E5+%2B-+2%283p%29%5E5\"\"%22%2B%22\"\"5%283p%29%5E4+%2B-+20p%283p%29%5E2\"\"%22%2B%22\"\"5%283p%29%5E2%281%29%5E4+%2B-+2%283p%29\"\"%22%5D%22\" \"%22%2B+1%22\" .
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