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### Sum of Integers

```
Date: 07/03/2001 at 03:57:14
From: Mark
Subject: Sum of integers

How many integers are 13 times the sum of their digits?
```

```
Date: 07/03/2001 at 13:12:33
From: Doctor Rob
Subject: Re: Sum of integers

Thanks for writing to Ask Dr. Math, Mark.

Suppose the number has four or more digits. Then

13*(a+b+c+d+...) = a + 10*b + 100*c + 1000*d + ...,
12*a + 3*b = 87*c + 987*d + ...,
4*a + b = 29*c + 329*d + ....

Now all the digits a, b, c, d, ..., are in the set
{0,1,2,...,9}, so

45 = 4*9 + 9 >= 4*a + b = 29*c + 329*d + ... >= 0.

This implies that d and all higher-order digits must be zero, and
c = 0 or 1.  If c = 0, then both a and b must also be 0, a
contradiction (or a solution, if you allow zero). Thus we have to
solve

4*a + b = 29,

with integers a and b between 0 and 9 inclusive.

This I leave to you.

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 07/03/2001 at 13:14:13
From: Doctor Ian
Subject: Re: Sum of integers

Hi Mark,

I'm going to assume that we can ignore negative integers, since the
sum of the digits of a negative integer is the same as the sum of the
digits of the corresponding positive integer.

For a one-digit integer, 'a', the condition is

a = 13(a)

This is true for a = 0, but not for any other digit.

For a two-digit integer 'ab', the condition is

10a + b = 13(a + b)

10a + b = 13a + 13b

0 = 3a + 12b

This is only true when a = 0 and b = 0.

For a three-digit integer 'abc', the condition is

100a + 10b + c = 13(a + b + c)

= 13a + 13b + 13c

0 = -87a + 3b + 12c

Again, this is true when a = 0, b = 0, and c = 0, and not for any
other set of digits.

A reasonable guess would be that 0 is the only number that satisfies
the condition, but we haven't proved anything. If you want to try to
prove it, proof by induction would probably be the way to proceed.

Does this help?

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

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