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/