Each letter must represent a digit from 1 to 9 with no
two letters representing the same digit.
Solution
We have: 9 x HATBOX = 4 x BOXHAT
set "HAT" =x and "BOX"=y(x,y has
3-digit)
the equation becomes
9(1000x+y) = 4(1000y+x) or
8996x = 3991y
which on division by 13 becomes
692x = 307y
where the coefficients are relatively prime.
This has the obvious solution we can set as:
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
Thus: H=3,A=0,T=7,B=6,O=9,X=2
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