
Re: A Good Activity
Posted:
Dec 7, 2012 8:02 AM


On Dec 6, 2012, at 1:15 PM, Louis Talman <talmanl@gmail.com> wrote:
> >> ... 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 sixdigit numbers, n, the rule is this: n is divisible by 7 if, and only if, the sum of the onesdigit, three times the tensdigit, two times the hundredsdigit, six times the thousandsdigit, four times the tenthousandsdigit, and five times the hundredthousandsdigit is divisible by seven. That is > > units + 3 tens + 2 hundreds + 6 thousands + 4 tenthousands + 5 hundredthousands. > > 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 tenthousands + 4 hundredthousands. > > 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>
Thanks. The development of the 3rule you show is pretty good. Given a number "abc" (a,b,c being digits) then expand it to 100a+10b+c and then rewrite it as 99a+9b+a+b+c and then see that for "abc" to be divisible by 3 (or 9), a+b+c must be divisible by 3 (or 9), since "99a+9b" is already divisible by 3 (or 9). Of course, another aspect is to see that this can be extended indefinitely for any number of digits (9,99,999,9999,...). I like it, thanks.
He isn't quite there yet, but soon. I can show him the progression in the sum of the digits as we add 1 to the number each time.
Bob Hansen

