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9 x HATBOX = 4 x BOXHAT

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Date: 05/30/2000 at 23:46:34
From: Stephen H. Stribling
Subject: HATBOX or is it BOXHAT

I've worked on this problem for 31 years and have even given it to
several people to answer. I gave it to a professor at a university and
he said, "There is no solution." I was told when I first received the
problem that there is definitely an answer. Here is the problem.

9 x HATBOX = 4 x BOXHAT

Each letter must represent a digit from 1 to 9 with no two letters
representing the same digit. 0 is not allowed. The "H" in HATBOX must
be the same as the "H" in BOXHAT and so on.

I hope you will be able to solve it.

Thanks,
Steve
```

```
Date: 05/31/2000 at 09:09:49
From: Doctor Peterson
Subject: Re: HATBOX or is it BOXHAT

Hi, Stephen.

Sorry - it's ALMOST solvable, but misses on one detail. Maybe you
changed the statement of the problem slightly from what you were
originally given.

If we call the number "HAT" x and "BOX" y, the equation becomes

9(1000x+y) = 4(1000y+x)

which simplifies to

8996x = 3991y

which on division by 13 becomes

692x = 307y

where the coefficients are relatively prime.

This has the obvious solution

x = 307n, y = 692n for any integer n

The only solution in which x and y are both 3-digit numbers is for
n = 1. Then

HAT = 307
BOX = 692

9(HATBOX) = 9(307692) = 2769228
4(BOXHAT) = 4(692307) = 2769228

The only thing wrong is that A = 0, which violates the conditions.
Since this is the only solution to the equation, there is no solution
to the problem as stated.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra
High School Puzzles

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