Associated Topics || Dr. Math Home || Search Dr. Math

### Sum of Digits Divisible by 11

```
Date: 08/16/99 at 08:05:26
From: Anoop
Subject: Divisibility

Dear Sir,

I've been stuck on this question for days now. Prove that in a
sequence of 39 consecutive natural numbers there exists at least one
number such that the sum of its digits is divisible by 11.

Please help me out. Even my math teacher is stumped.
Anoop.
```

```
Date: 08/16/99 at 13:38:21
From: Doctor Rob
Subject: Re: Divisibility

Thanks for writing to Ask Dr. Math.

The digital sum of the successor of a natural number is 1 greater than
the digital sum of the number, except when the number ends in 9. When
a number ends in 9, the digital sum of its successor is 8 less than
the digital sum of the number, except when the number ends in 99. When
a number ends in 99, the digital sum of its successor is 17 less than
the digital sum of the number, except when the number ends in 999.

The general rule is that if a number X ends with exactly N 9's, then
the digital sum of X+1 minus the digital sum of X equals 1 - 9*N.

Suppose you did have 38 consecutive numbers A, A+1, ..., A+37 all of
whose digital sums were not divisible by 11. Then you can prove that
both A-1 and A+38 have digital sums divisible by 11, so there aren't
39 numbers in a row with that property.

Start by realizing that if none of 38 consecutive numbers ends with
99, then one of their digital sums must be divisible by 11, since
starting with a number that ends in a 1, the pattern in digital sums
goes:

Final Digit
0    1    2    3    4    5    6    7    8    9
-------------------------------------------------
c  , c+1, c+2, c+3, c+4, c+5, c+6, c+7, c+8,
c  , c+1, c+2, c+3, c+4, c+5, c+6, c+7, c+8, c+9,
c+1, c+2, c+3, c+4, c+5, c+6, c+7, c+8, c+9, c+10,
c+2, c+3, c+4, c+5, c+6, c+7, c+8, c+9, c+10,

So eleven consecutive values of the digital sum are covered by at
least one of any 29 consecutive numbers, and one of those eleven
consecutive values must be divisible by 11. Thus one of them must end
in 99.

Next see that A+18 should be the number ending in 99.

Then see that it cannot end in exactly 2, 3, 4, or 5 nines, for a
similar reason. (Make a similar chart for each.)

Finally, show that A+18 can end with exactly 6 nines, and must have
digital sum of the form c = 11*k + 10.  Then A-1 and A+38 have digital
sums 11*k and 11*(k-3), respectively.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Number Theory

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search