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### Money Puzzle

```
Date: 01/14/98 at 08:37:29
From: Vedachal
Subject: Puzzle

Hi Dr. Math,

Can you help me out with this puzzle?

A man goes to the bank and asks for x dollars and y cents. The
banker by mistake gives him y dollars and x cents. After spending
5 cents the man realises that he now has twice the amount he asked
for.  What was the amount he asked for?

Regards,
Veda
```

```
Date: 01/14/98 at 09:56:58
From: Doctor Rob
Subject: Re: Puzzle

Thanks for writing, Veda.

The amount he asked for was 100*x + y cents. The amount he received
was 100*y + x cents. After spending five cents, he was left with
100*y + x - 5 cents.  Then

100*y + x - 5 = 2*(100*x + y),
98*y - 199*x = 5.

You want to find a solution to this equation for which both x and y
are nonnegative integers, since the problem implies that that must be
so. The problem also implies that x < 100 and y < 100.

Equations requiring integer solutions are called Diophantine
equations. This one is first-degree in x and y, so it is a linear
Diophantine equation. The following technique works.

We want to reduce the size of the coefficients of the unknowns
x and y. We start by replacing the larger one as follows.
Since 199 = 2*98 + 3, we can rewrite the equation as

98*y - (2*98 + 3)*x = 5
98*(y - 2*x) - 3*x = 5

Put z = y - 2*x, which will still be an integer.  Then y = z + 2*x.
Then

98*z - 3*x = 5

Now since 98 = 32*3 + 2, we can rewrite this equation as

(32*3 + 2)*z - 3*x = 5
2*z - 3*(x - 32*z) = 5

Put w = x - 32*z, which will still be an integer.  Then x = w + 32*z.
Then

2*z - 3*w = 5

Now since 3 = 1*2 + 1, we can rewrite this equation as

2*z - (1*2 + 1)*w = 5
2*(z - w) - w = 5

Put v = z - w, which will still be an integer.  Then z = v + w.  Then

2*v - w = 5
w = 2*v - 5

Now if v is any integer, we can work backwards to figure out all the
other variables in terms of v.  We have

w = 2*v - 5
z = v + w = v + (2*v - 5) = 3*v - 5
x = w + 32*z = (2*v - 5) + 32*(3*v - 5) = 98*v - 165
y = z + 2*x = (3*v - 5) + 2*(98*v - 165) = 199*v - 335

These last two equations give all solutions to the equation we started
with, 98*y - 199*x = 5. We still must use the conditions on x and y.
Now in order that

100 > x >= 0
100 > 98*v - 165 >= 0
265 > 98*v >= 165
265/98 > v >= 165/98

Similarly,

100 > y >= 0
100 > 199*v - 335 >= 0
435 > 199*v >= 335
435/199 > v >= 335/199

You can figure out what the only possible integer value of v is, and
that will tell you what x and y are, once you know v.

Similar problems can be handled in the same kind of way, by reducing
the coefficients of the unknowns (y, x, z, w, v in this case, reduced
via 198, 99, 3, 2, and 1) until one of them divides the other evenly
(in this case, 1 divides 2 evenly), then working backwards.

-Doctor Rob,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Basic Algebra
High School Linear Equations
High School Puzzles

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