Divisibility by 11

By Christopher Nicolson on Monday, August 19, 2002 - 09:07 pm:

About a year ago I discovered something that I couldn't explain whilst mucking about during my lunch break last summer.

If you have a number and you subtract the lead digit into the subsequent digit and that answer into the next digit, and so on and you end up with a total of 0 at the end of the number then it is divisible by 11

This sounds slightly proposterous, but consider 143 (we know this is divisble by 11)

4-1=3
3-3=0

Can anybody explain this, is it wrong or what?

It has been bugging me for a while

Cheers,
Christopher

By Paul Smith on Monday, August 19, 2002 - 10:46 pm:

This is true, although there are questions of necessity and sufficiency.

Consider a number

n

written

n = " a_{m} a_{m - 1} \dots a_{0}''

. What you are saying is that if

a_{0} - (a_{1} - (\dots (a_{m - 1} - a_{m}) \dots)) = 0

then

n

is divisible by 11.

Firstly we can simplify this condition to give $a_{0} - (a_{1} - (\dots (a_{m - 1} - a_{m}) \dots)) = a_{0} - a_{1} + \dots + (- 1)^{m} a_{m}$ . Now we note that $10 \equiv - 1 (\mod 11) \Rightarrow 10^{i} \equiv (- 1)^{i} (\mod 11)$ , where $i$ is an integer. Since $n = 10^{0} a_{0} + 10^{1} a_{1} + \dots + 10^{m} a_{m}$ , we have $n \equiv a_{0} - a_{1} + \dots + (- 1)^{m} a_{m} (\mod 11)$ , and hence $n \equiv a_{0} - (a_{1} - (\dots (a_{m - 1} - a_{m}) \dots)) (\mod 11)$ . Thus $n$ is divisible by 11 if $a_{0} - (a_{1} - (\dots (a_{m - 1} - a_{m}) \dots)) = 0$ , but also if it equals 11 (e.g. 902) or 22 (e.g. 90904) or ..., and therefore your condition is sufficient, but not necessary for divisibility by 11.
Hope this helps.

Paul

By Julian Pulman on Monday, August 19, 2002 - 11:01 pm:

Well, say we have a number

N

, where the first digit is

d_{1}

, and generally the

n^{th}

digit is

d_{n}

(e.g.

1532 \Rightarrow d_{1} = 2

d_{2} = 3

d_{3} = 5

d_{4} = 1

) then:

$N = d_{1} + 10 d_{2} + 10^{2} d_{3} + \dots + 10^{n} d_{n + 1}$

Since $10 \equiv - 1 (\mod 11)$ ,

$N \equiv d_{1} - d_{2} + d_{3} + \dots - (- 1)^{n} d_{n} (\mod 11)$

Clearly, it is a sufficient condition for divisibility by 11 if the difference and sum of successive terms has zero magnitude, since all numbers divide 0. Furthermore, we require only that the sum of the odd digits minus the sum of the even digits be a multiple of 11.

Clearly, your example works because the sum of odd digits is 1+3 = 4, and the sum of even is 4, their difference is zero. Consider 16060418, and observe (1 + 0 + 0 + 1) - (6 + 6 + 4 + 8) = -22 = -2x11
Hence, 16060418 is divisible by 11.
(11x1460038)

Julian