```Date: Dec 6, 2012 1:15 PM
Author: Louis Talman
Subject: Re: A Good Activity

> ... Of course it would be a number involving 7 and 13. We didn't have a  > rule for 7 and 13 was outside of his multiplication facts.>It's not hard to develop a rule for seven. Maybe he can do it if you first  ask him why the rule for 3 works, and then show him if he can't figure  that out himself. [Hint: 725 = 7(100) + 2(10) + 5 = 7(99 + 1) + 2(9 + 1) +  5.]For six-digit numbers, n, the rule is this:  n is divisible by 7 if, and  only if, the sum of the ones-digit, three times the tens-digit, two times  the hundreds-digit, six times the thousands-digit, four times the  ten-thousands-digit, and five times the hundred-thousands-digit is  divisible by seven.  That is  units + 3 tens + 2 hundreds + 6 thousands + 4 ten-thousands + 5  hundred-thousands.For numbers of more than six digits, use the fact that the coefficients  are periodic, with period of length six.Divisors greater than ten are a little more difficult, because they're  most efficiently done using negative numbers. For example, a number is  divisible by eleven iff the sum of its digits *with alternating sign* is  divisible by eleven.For thirteen, the appropriate sum is  units - 3 tens - 4 hundreds - thousands + 3 ten-thousands + 4  hundred-thousands.The coefficients are again periodic, and the length of the period is again  six.  (Coincidence???  !!)The rules can be generalized to arbitrary bases. Thus, for example, a  number written in hexadecimal is divisible by fifteen iff and only if the  sum of its hexadecimal digits is divisible by fifteen.So another possibility for divisibility by seven is to write the number in  octal. Then the number is divisible by seven iff the sum of its octal  digits is divisible by seven.- --Lou Talman   Department of Mathematical & Computer Sciences   Metropolitan State University of Denver<http://rowdy.msudenver.edu/~talmanl>
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