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

### Determine b if (a+b)/5 = (b-1)/2

```Date: 07/25/2003 at 01:09:20
From: M.Vijay Kumar
Subject: Number Theory

Both a and b are 4 four-digit numbers, a>b, and one is obtained from
the other by reversing the digits. Determine b if  (a+b)/5 = (b-1)/2.
```

```
Date: 07/26/2003 at 11:23:11
From: Doctor Greenie
Subject: Re: Number Theory

Many thanks for sending us this problem - I got a lot of good mental
exercise on it. Please note that a much more elegant solution may be
possible; this is simply the method I came up with that works.

We are given that

(a+b)/5 = (b-1)/2

Working with this equation, we have

2(a+b) = 5(b-1)
2a+2b = 5b-5

or

(1)  3b-2a = 5

From this we can conclude that

(2)  a = 1.5b (approximately)

If we represent the smaller 4-digit number (b) by the string pxyq,
then the larger 4-digit number (a) is the string qyxp. This equation
(2) places restrictions on the combinations of first and last digits p
and q that our numbers a and b can have. The following table lists all
the possible 4-digit combinations for the numbers a and b based on the
restrictions imposed by equation (2) above:

b      a
------------
1xy2    2yx1
2xy3    3yx2
2xy4    4yx2
3xy4    4yx3
3xy5    5yx3
4xy6    6yx4
4xy7    7yx4
5xy7    7yx5
5xy8    8yx5
6xy9    9yx6

For any of the above possibilities, representing the first and last
digits of the number b by p and q, respectively, we have, from (1)
above), the following:

3b - 2a = 5

or

3(1000p+100x+10y+q)-2(1000q+100y+10x+p) = 5
300x+30y+3(1000p+q)-200y-20x-2(1000q+p) = 5

which we can rearrange as

(3)  (280x-170y)+3000p-2000q = 2p-3q+5

The variables p, q, x, and y are all single-digit positive integers,
so the left-hand side of this equation is divisible by 10.  The right-
hand side of the equation must therefore also be divisible by 10.

The table above shows all the possible combinations of digits p and q;
let's find which of those combinations satisfy the condition that

2p-3q+5

is divisible by 10:

p   q   2p-3q+5
-----------------
1   2   2-6+5 = 1
2   3   4-9+5 = 0
2   4   4-12+5 = -3
3   4   6-12+5 = -1
3   5   6-15+5 = -4
4   6   8-18+5 = -5
4   7   8-21+5 = -8
5   7   10-21+5 = -6
5   8   10-24+5 = -9
6   9   12-27+5 = -10

So now there are only two possibilities for the digits p and q:

(p,q) = (2,3)
or
(p,q) = (6,9)

Now let's try each of these two possibilities in equation (3) above,
remembering that p, q, x, and y are single-digit integers.

(280x-170y)+3000p-2000q = 2p-3q+5

For (p,q) = (2,3) we find

280x-170y+6000-6000 = 4-9+5 = 0
280x-170y = 0
28x-17y = 0

The only solution in single-digit integers for this equation is
(x,y) = (0,0); this gives us the solution to the problem p=2, q=3,
x=y=0.

Checking this solution, we have

b = 2003
a = 3002

(a+b)/5 = 5005/5 = 1001
(b-1)/2 = 2002/2 = 1001

For (p,q) = (6,9) we find

280x-170y+18000-18000 = 12-27+5 = -10
280x-170y = -10
28x-17y = -1

We now solve this as a diophantine equation with x and y single-digit
integers:

17y = 28x+1
17y = 17x+11x+1
y = x + (11x+1)/17

The expression (11x+1) must be divisible by 17; trying single-digit
integer values for x, we have

11(1)+1 = 12
11(2)+1 = 23
11(3)+1 = 34

34 is divisible by 17.  So we now have

x = 3
y = 3 + (34/17) = 3+2 = 5

And we have a second solution to the problem:  p=6, x=3, y=5, q=9.

b = 6359
a = 9536

(a+b)/5 = 15895/5 = 3179
(b-1)/2 =6358/2 = 3179

So finally we have two solutions to the problem:

(a,b) = (3002,2003)

and

(a,b) = (9536,6359)

I hope you can follow all of this.  Please write back if you have any
further questions about any of the work I have done.

And thanks again for a great problem!

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College 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