How can verify a number( say 4 or more digits) divisible by 11 ?

Answer 1

An integer number is divisible by 11 if and only if the sum of all its odd digits (first, third, ...) is different from the sum of the sum of all even digits (second, fourth...) from a multiple of 11

1 908 071 is multiple of 11
because 1+ 0+ 0+1 = 2
and 9+8 +7 = 24
finally, 24 - 2 = 22 = 2 * 11

2 981 also
because 2 +8 = 10
and 9+1 = 10
finally, 10 - 10 = 0 = 0 * 11

1 995 no
because 1 +9 = 10
and 9+5 = 14
finally, 14 - 10 = 4 not multiple of 11

Answer 2

Test for divisibility by 11. Subtract the last digit from the remaining leading truncated number. If the result is divisible by 11, then so was the first number. Apply this rule over and over again as necessary.
Example: 19151--> 1915-1 =1914 -->191-4=187 -->18-7=11, so yes, 19151 is divisible by 11.

Answer 3

Is there a less cumblesome way to looking at this (Yes, below):

Labeling from the left, let digits of n be d1, d2, ..., dk.
i.e. n = d1x10^(k-1) + d2x10^(k-2) + ... + dkx10^0
Since 11n = 10n + n, 11n is the addition of n and a left shifted n.
So, the addition 10n+n is adding every 2 consecutive digits in n together, d1+d2, d2+d3, ..., d(k-1)+dk, except the extreme left and right where d1 and dk only added once.

Consider cases where all these additions were less than 10.
The digits in 11n would be d1, d1+d2, d2+d3, ..., d(k-2)+d(k-1), d(k-1)+dk, dk
Sum of odd digits in 11n = d1 + (d2 + d3) + (d4+d5) + ... a,
where a = d(k-1)+dk if k is odd, and a = dk if k is even.
Sum of even digits in 11n = (d1 + d2) + (d3 + d4) + (d5 + d6) + ... b,
where b = d(k-1)+dk if k is even, and b = dk if k is odd.
Thus, these two sums are equal, so their difference is 0, which is divisible by 11.

Now, lets consider if some of a sum of 2 consecutive digits is > 10, then due to the carry forward in 10n+n, the corresponding sum of these 2 digits becomes di+d(i+1)-10, while the sum of digits left of this pair becomes d(i-1)+di+1, resulting in a decrement of 10 and an increment of 1 to the sum of the odd and even (or even and odd) digits, respectively. Thus the difference of the two sums will be increased by 10-(-1)=11, or 1-(-10)=-11 from the one without any carry over. Thus, this difference between sums of odd and even digits is still divisible by 11.

Thus, the changes to the difference of the sums of odd and even digits due to multiple di+d(i+1)>10 are addition or subtraction of multiples of 11 to the direct sums of di+d(i+1) without doing carry forward.

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jdsheth2004's posting of Savory's site led me to wonder about his method, and figure out a way to reason out his method:

To check the division of any odd numbers:
1. Find out the first multiple that ends with digit '1' or '9'.
2. Note the number, m = round(this multiple/10).
3. Thus, the checking method is either
3a. if end with 1, truncate the last digit away and subtract the product of it and m from the number formed by the remaining digits.
3b. if end with 9, truncate the last digit away and add the product of it and m from the number formed by the remaining digits.

Mathematically (all variables are integers), to check divisibility by q of n:
If m' = 10 x m and (m'+1) = q x r,
then (n - (m'+1)) = n - q x r, thus n is divisible by q if n-q x r is divisible by q.
If last digit of n is dk,
then n - dk x q x r
= n - dk (m'+1)
= n - (dk x m') - dk
= 10 ((n - dk)/10 - (dk x m)),
where (n-dk)/10 is the number after truncating last digit away.
If (n - dk)/10 - (dk x m) divisibile by q,
then 10 ((n - dk)/10 - (dk x m)) also divisible by q,
then n - dk x q r also divisible by q,
thus n is divisible by q.
(For ending with 9, it m = (m'-1))

For the case of 13, multiples are 13, 26, 39, thus use 39.
So, m' = 40, m = 4, and the method is truncate away last digit, dk, and add 4 x dk to the remaining.

And using this way, for 11, after subtracting till the second digit, we will get the same formula, since
d1 - (d2 - (d3 - (d4 - (....d(k-1) - dk))..)))) + s(1+10)
= d1 - d2 + d3 -d4 + ... dk + 11s
= difference between odd and even numbers (ignoring the sign) + 11s
where s is the number of times dk < d(k+1)-(d(k+2)-(...)) which will cause a borrowing from dk.

Answer 4

if x mod 11 = 0 it is divisible by 11 where x = the number you want to verify.

Answer 5

Calculator -- (4 digit number) divided by 11

Answer 6

Just check the calculator, unless you want some really long division to do

Answer 7

Try dividing it.