2006-08-01 04:46:53 · 8 answers · asked by MARTA SUSANA L 3 in Science & Mathematics ➔ Mathematics
From about.com:
Dividing by 7 (2 Tests)
* Take the last digit in a number.
* Double and subtract the last digit in your number from the rest of the digits.
* Repeat the process for larger numbers.
For example: 357
- Double the 7 to get 14. Subtract 14 from 35 to get 21. 21 is divisible by 7, which means that 357 is divisible by 7.
NEXT TEST
* Take the number and multiply each digit beginning on the right hand side (ones) by 1, 3, 2, 6, 4, 5. Repeat this sequence as necessary
* Add the products.
* If the sum is divisible by 7 - so is your number.
For example, 2016:
6(1) + 1(3) + 0(2) + 2(6) = 21
Since 21 is divisible by 7, so is 2016.
For more info, check out:
http://math.about.com/library/bldivide.htm
2006-08-01 04:53:07 · answer #1 · answered by Josh 2 · 2⤊ 0⤋
Based on the fact that (a * 10 + b) mod 7 = (a - 2 * b) mod 7, you keep multiplying the last digit of the number by 2 and subtract it from the rest of the digits in the number. If you finally get a number dividable by 7, the number is dividable by 7.
Example: 8876
887 - 2 * 6 = 875
87 - 5 * 2 = 77
77 is dividable by 7. So 8876 is divisible by 7.
2006-08-01 05:12:07 · answer #2 · answered by Stanyan 3 · 0⤊ 0⤋
Taken from Wikipedia:
Sum the digits in alternate blocks of three from right to left, then difference the two sums.
Sum the number with the last two digits removed, doubled, plus the last two digits.
Sum the number with the last digit removed with 5 times the last digit.
Difference the number with the last digit removed with 2 times the last digit.
There are others for non base-10 mathematics.
2006-08-01 10:12:55 · answer #3 · answered by gestaltdream 1 · 0⤊ 0⤋
I *really* like the double and subtract method mentioned above! The main difficulty is that it doesn't preserve the remainder after division by 7, although it does preserve divisibility. Very cute!
A similar method for 13 would be to multiply the last digit by 4 and add to the others.
2006-08-01 05:05:57 · answer #4 · answered by mathematician 7 · 0⤊ 0⤋
7 is a prime number which is divided by 7 and 1
2006-08-01 05:02:25 · answer #5 · answered by SAMUEL D 7 · 0⤊ 0⤋
The rule is so bizarre, it's easier just to divide by 7 to see if there's a remainder.
2006-08-01 04:51:57 · answer #6 · answered by MollyMAM 6 · 0⤊ 0⤋
there is no easy rule for that one... (not like 3 or 2 or 11...)
just try to divide it, and you'll know. That's the rule, for that one.
2006-08-01 04:50:38 · answer #7 · answered by Anonymous · 0⤊ 0⤋
the divisibility rule makes it more complicated than the trial and error method
2006-08-01 04:50:49 · answer #8 · answered by raj 7 · 0⤊ 0⤋