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### A Variable for Each Digit

```
Date: 11/28/2001 at 02:26:53
From: Joe Smith
Subject: Inverse ages

A husband's age (h) is the inverse of his wife's age (w). His age is
also greater than her age. Also 11(h-w) = h+w. Some example ages are
82 and 28 or 54 and 45. Is there a way to solve this without simple
trial and error?

I have tried to establish some relation for inverse numbers, but the
only thing I know of is the divisible by 11 thing.
```

```
Date: 11/28/2001 at 12:41:21
From: Doctor Peterson
Subject: Re: Inverse ages

Hi, Joe.

Apparently you are using the term "inverse" to mean "reversal of the
digits"; this is not the usual meaning of the term. The best way to
work with problems involving digits is to use a separate variable for
each digit, so that

h = "ab" = 10a + b
and
w = "ba" = 10b + a

Now the sum and difference are

h+w = 10(a+b) + (a+b) = 11(a+b)

h-w = 10(a-b) + (b-a) = 9(a-b)

(Notice what this tells us about divisibility of the sum and
difference of reversed 2-digit numbers.)

You are told that 11(h-w) = h+w, which means that

11[9(a-b)] = 11(a+b)

which simplifies to

99a - 99b = 11a + 11b

88a = 110b

44a = 55b

4a = 5b

Since a and b are both single digits, and we see that a must be a
multiple of 5 and b must be a multiple of 4, the only solutions of
this are

a = 0, b = 0
a = 5, b = 4

The first of these is not allowed, so we have

h = "ab" = 54
w = "ba" = 45

11(h-w) = 11(9) = 99
h+w = 99

So this works.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Algebra
Middle School Word Problems

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