Theorem

Divisibility by $11$ from the alternating digit sum

Arithmetic • Number System • Divisibility

Formula

d_{n} \dots d_{1} d_{0} \equiv d_{0} - d_{1} + d_{2} - \dots (mod 11)

A whole number is congruent modulo

11

to its alternating digit sum, formed from the units digit with sign

+

and alternating leftwards; so it is divisible by

11

exactly when that alternating sum is a multiple of

11

, zero included.

How to use it

1Add the digits in the odd places and the digits in the even places separately, then subtract: it is the difference that must be a multiple of $11$ .
2Combine it with the digit-sum test for $3$ to settle divisibility by $33$ , and with the test for $9$ for $99$ .
3On a digit pattern with unknowns the alternating sum usually collapses to a single digit, and a single digit is a multiple of $11$ only when it is $0$ .

Worked example

918082

divisible by

11

Step-by-step

1
$2 - 8 + 0 - 8 + 1 - 9 = - 22$
Why: Alternate the signs starting from the units digit and working leftwards.
2
$- 22 = 11 \times (- 2)$
Why: A multiple of $11$ , negative or not, means the number itself is divisible by $11$ .

Answer

Yes:

918082 = 11 \times 83462

Watch out

The plain digit sum tests

3

and

9

and says nothing about

11

- the signs must alternate. Starting the alternation at the leading digit instead of the units digit flips the sign of the whole sum, which is harmless for a divisibility verdict but wrong if the residue itself is wanted.

Also known as

divisibility by 11, alternating sum of digits, mod 11 digit test, odd and even place digits, test for 11, divisibility rule for 11

Divisibility by $11$ from the alternating digit sum

Formula

Worked example

Appears in past papers

2018 (II)