How do you know if a number is dividable by 7? and 11?

Question

I already know how to decide if a number devides by 2,3,4,5,6,8 and 9. I only need 7 and 11.
Is there a better way to know if a number is divideable by 8 other then the last 3 digits?

Thank you ;)

Answer 1

To find out if a number is divisible by seven, take the last digit, double it, and subtract it from the rest of the number.
If you get an answer divisible by 7 (including zero), then the original number is divisible by seven. If you don't know the new number's divisibility, you can apply the rule again.
Check to see if 203 is divisible by 7.

double the last digit: 3 x 2 = 6
subtract that from the rest of the number: 20 - 6 = 14.
check to see if the difference is divisible by 7: 14 is divisible by 7, therefore 203 is also divisible by 7.

For 11:

Take the first digit, subtract the second digit from it, add the third digit, subtract the fourth digit and so on. If the number is divisible by 11, the answer will be zero.
consider the number is 12354.
Do this calculation:
1-2+3-5+4
Answer is 1. This is not equal to 0 thus the number is not divisible by 11.

Consider 108416:

1-0+8-4+1-6=0
Thus the number is divisible by 11!

Answer 2

divisibility by 7:

double the last digit of the number and subtract it to the remaining number if the difference is divisible by 7 then its divisible by 7. if the number is still large continue until you can say it is divisible or not

e.g

105 --- 5*2=10 --- 10-10=0 --- 0 is divisible by 7 therefore 105 is divisible by 7

275 --- 5*2=10 --- 27-10=17 --- 17 is not divisible by 7 therefore 275 is not divisible by 7

divisibility by 11:

add the odd order digits and the even ordered digits subtract the sums if it the difference is divisible by 11 then the number is divisible by 7

e.g.

12345 --- 1+3+5=9, 2+4=8 --- 9-8=1 --- 1 is not divisible by 11 therefore the number is not divisible by 11

1023 --- 1+2=3, 0+3=3 --- 3-3=0 --- 0 is divisible by 11 therefore the number is divisible by 11

Answer 3

well for eleven u just add up consecutive digits and the diferance should be either 0 or a multiple of 11
e.g 596607
step1) 5+6+0=11
step2) 9+6+7=22
step3) 22-11=11 , thus 596607 is divisable by 11

im not sure about the divisibilty test for 7. and the test for 8 is what u say , i dont know anything else for 8.

Answer 4

1 mod 7=1
10 mod 7=3
100 mod 7=2
1000 mod 7= -1
10000 mod 7= -3
100000 mod 7= -2
1000000 mod 7=1 and the cycle repeats...

So add up the unique, 3 times tens, 2 times hundreds, ... , of the main number together and if the arithmetic sum is dividable by 7 then the main number is too.
Example: A=356975208
8x1 + 0x3 + 2x2 + 5x(-1) + 7x(-3) + 9x(-2) + 6x1 + 5x3 + 3x2
= -5
so is not dividable and remains 2 (A=50996458x7 + 2)
Example: A=35462
2x1 + 6x3 + 4x2 + 5x(-1) + 3x(-3) = 14
so it is dividable (A=5066x7)

1 mod 11=1
10 mod 11= -1
100 mod 11= 1
1000 mod 11= -1 and the cycle repeats...

So add up all digits of the number reversing its sign interchanged; then if the arithmetic sum is dividable by 11 then the main number is too.
Example:
A=232185325
5-2+3-5+8-1+2-3+2=9
so it is not dividable by 11 and remains 9 (A=21107756x11+9)
Example:
A=684211
1-1+2-4+8-6=0
so it is dividable by 11 (A=62201x11)

Answer 5

I know for 11 but 7 is still missing. It's very easy and works every time. Just combine the first and last number and if you get the middle numbers it is divideable

Answer 6

I only know about the numbers divides by 2

Answer 7

11 :

The numbers have to be like so: 11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143, 154.......And so on..

7: Sorry, my math teacher came up with one, and I forgot.
Hope someone can help you with this.

Answer 8

For 7, try this link (http://polymathematics.typepad.com/polymath/2005/08/divisibility_te.html )

and for 11
(http://www.sonoma.edu/users/w/wilsonst/Courses/Math_300/fractions/dc.html )

Answer 9

This site has divisibility rules:

http://argyll.epsb.ca/jreed/math7/strand1/1104.htm

Answer 10

anything you can mulitply them by, you can divide by.