9 x HATBOX = 4 x BOXHAT

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/