9 x HATBOX = 4 x BOXHATDate: 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/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/