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### No-Solution Equations

```Date: 11/09/2002 at 16:23:25
From: Mussy
Subject: No-solution equations

Hi,

Here's the question:

For any positive integer n, let S(n) denote the sum of its digits.
Show that the equation n + S(n) = 1,000,000 has no solution. Then
solve the equation n + S(n) = 1,000,000,000.

Thank so much.
Mussy
```

```
Date: 11/10/2002 at 13:07:05
From: Doctor Greenie
Subject: Re: No-solution equations

Hi, Mussy -

If we have

n + S(n) = 1,000,000

then n is a 6-digit number; this means S(n) is at most 6(9) = 54.
Therefore, n must be a number of the form

n = 999,9AB

where the digits A and B are to be determined. Rewriting our 6-digit
number 999,9AB as 999,900 plus the two-digit number AB, we then have

n + S(n) = 1,000,000

(999,900 + 10A + B) + (4(9) + A + B) = 1,000,000

999,936 + 11A + 2B = 1,000,000

11A + 2B = 64

This equation has no solution (A,B) with both A and B being single-
digit integers. (A=4 would make B = 10; A = 5 would make B not an
integer; A = 6 would make B negative.)

So there is no solution to the equation

n + S(n) = 1,000,000

On the other hand, if we have

n + S(n) = 1,000,000,000

then n is a 9-digit number; this means S(n) is at most 9(9) = 81.
Therefore, n must be a number of the form

n = 999,999,9AB

where the digits A and B are to be determined. Rewriting our 9-digit
number 999,999,9AB as 999,999,900 plus the two-digit number AB, we
then have

n + S(n) = 1,000,000,000

(999,999,900 + 10A + B) + (7(9) + A + B) = 1,000,000,000

999,999,963 + 11A + 2B = 1,000,000,000

11A + 2B = 37

This equation does have a solution with A and B both single-digit
positive integers: A = 3 and B = 2. So the solution for this case is

n = 999,999,932

I hope this helps. Please write back if you have any further

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

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