Divisibility?

Question

How do you check the divisibility of a number by 7,9,11,13,17? Answer if you know atleast one of them. (Ex: Divisiblity by two the no. should end with an even no.)

Answer 1

Tests for 9 you probably have in abundance here.

For 7, take the last digit off of a number, double it, and subtract it from the rest... if the smaller number is divisible by 7, then the original one is, too.

Example: Is 444318 divisible by 7?
Take off the 8, double it, subtract from the rest:
44431 - 16 = 44415
Keep doing this 'til you find out...
4441 - 10 = 4431
443 - 2 = 441
44 - 2 = 42. Yup. 444318 is divisible by 7, because 42 is divisible by 7.

For 17, use the same method as for 7, except multiply the chopped-off digit by 5 instead of doubling it.

Example: Is 6997591 divisible by 17?
699759 - 5 = 699754
69975 - 20 = 69955
6995 - 25 = 6970
697 - 0 = 697
69 - 35 = 34. Yup. 6997591 divisible by 17 because 34 is divisble by 17.

The reasons these work are, for 7 you're essentially striking out 21's, and for 17 you're striking out 51's. You could use a similar method for 13 by multiplying the chopped-off digit by 9 and subtracting from the rest, striking out 91's as you go. You could chop off 11 at a time like this for a divisibility test, but here's another way:

For 11, subtract the two sums of every other digit... if it's divisible by 11, the original number is.

Example: Is 4527853 divisible by 11?
4 + 2 + 8 + 3 = 17
5 + 7 + 5 = 17
17 - 17 = 0. Yup. 4527853 is divisible by 11, because 0 is divisble by 11.

Because 7 × 11 × 13 = 1001, here's a method you could use for all three: strike off multiples of 1001. It makes for very easy subtractions, and you'd be left with a three-digit number that would hold for all three divisors.

Answer 2

Divisibility by 9: Add up the decimal digits of the number. If the sum is divisible by 9, so is the original number. This can be repeated.
Ex: 3591->3+5+9+1=18 *is* divisible by 9.

Divisibility by 11: Alternately add and subtract the digits of the number. If the result is divisible by 11, so is the original number. This can be repeated.
Ex: 8723->8-7+2-3=0 *is* divisible by 11.

Divisibility by 7: Take the last digit, double it, and subtract from the rest of the number. If the result is divisible by 7, so is the original. This can be repeated.
Ex: 7398->739-16=723->72-6=66 is *not* divisible by 7.

Divisibility by 13: Take the last digit, multiply it by 4 and add to the rest of the number. If the result is divisible by 13, so is the original. This can be repeated.
Ex:5941->594+4=598->59+32=91->9+4=13 *is* divisible by 13.

Divisibility by 17: Take the last *two* digits, multiply them by 9. Either subtract this from the rest or subtract the rest from this.
Ex: 77129->771-261=510->90-5=85 *is* divisible by 17.

Answer 3

To check if any number is divisible by 9 or not then simply add the digits. if the result is 9 then the number is divisible by 9.

To check if any number is divisible by 11 you can add the alternative digits. The sum of both the sets should be the same.
For eg. for the number 1331. We add 1(1 should be the number of thousands place) and 3 (3 should be the tens place). Then the result is 4. Similarly, 1(of ones place) and 3 (of hundreds place) are added which results in 4. Therefore both the answers are 4. Hence, it is divisible by 11.

Answer 4

this is the trick ...

"you should calculate with the remainders"
for instance check if is divisible by 7
127 = 1*100 + 2 * 10 + 7 *1
1 / 7 remainder = 1
10 / 7 reminder = 3
100/7 remainder = 2

sooooo
127 = 1*100 + 2 * 10 + 7 *1 = (when divided by 7 ) =
127 = 1*2 + 2*3 + 7*1 = 2 + 6 + 7 = 2 + 6 + 0 = 1
( remember calculate with thr remainder when dividing by 7 )

so 127 is not divisible by 7 it will have remainder 1.

for 7 :
1 / 7 remainder = 1
10 / 7 reminder = 3
100/7 remainder = 2
1000/7 remainder = 6 = -1
10000/7 remainder = -3
100000/7 remainder = -9 = -2

93269

1*9 + 3*6 +2*2 -1*3 -3*9 = ... = ... if this is 0 then divisible by 7 otherwise the number is the remainder of the division.

You can make rules for any division

Answer 5

A number is divisible by 9 if the sum of its digits is divisible by 9.

Answer 6

I just divide them into a number and if there is no remainder then it is divisible. I don't know any easy trick. Of course calculators and computers help.

Answer 7

k.... with 7 ...i dont know.
9 - add each n every digits n if the no is divisible by 3 .. then it is divisibly by 9
11 - double no's (egs. 11 , 22, 33,, 44, 55,..............)not necessary

Answer 8

In general you use the arithmetic of remainders, called "modular arithmetic." You should Google it--it's very interesting and easy, and it's kind of fun to generate all of these rules yourself.

Answer 9

i know same nice way that i discover my self
but i wont tell... cuz i duno if its common