The divisibility by seven rule that Maria gave us today seems easy compared to this rule: I didn't try it yet, just reading it gives me a headache! Go to this link:
http://www.maa.org/mathland/mathtrek_05_23_05.html
For those of you who didn't take notes on this but are curious how it works, here is by 11.
Divisibility by 11:
Divisibility by 11 is the most interesting of the above tests (7 will be studied below). We do two sums (the odd numbered digits and the even numbered digits), subtract one sum from the other, and see if this is divisible by 11. By the way, if we end up with zero, then that is divisible by 11. We can repeat that process, just as we did with 3. Let's look at an example:
348719033+8+1+0=124+7+9+3=2323-12=11Is divisible by 11
We can, of course, do the summing in different orders. In fact we can just go from left to right adding and subtracting alternate digits: 3-4+8-7+1-9+0-3=-11 (divisible by 11).
Make sure you study the divisibility rules! They will be on the next test!
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Wow! That really is quite the algorithm for divisibility by 7. I think I'll just do the division. :)
ReplyDeleteI created a rule for divisibility by seven, eleven and thirteen whose algorithm for divisibility by seven is this:
ReplyDeleteN = a,bcd; a' ≣ ( − cd mod 7 + a ) mod 7; cd is eliminated and if 7|a'b then 7|N. The procedure is applied from right to left repetitively till the leftmost pair of digits is reached. If the leftmost pair is incomplete consider a = 0.
Example: N = 382,536, using simple language:
36 to 42 = 6; 6 + 2 − 7 = 1 → 15; 15 to 21 = 6; 6 + 3 − 7 = 2 → 28; 7|28 and 7|N.
This rule is mentioned in my unpublished (officially registered) book: Divisibility by 7, the end of a myth?.
It is applied in seconds to very large numbers entirely through mental calculation.