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Alphametric Problem


Date: 04/06/99 at 05:38:21
From: John Gracefield
Subject: Alphametric

Dr. Math,

Help me solve this alphametric, please, in which each letter 
represents a distinct digit.

                 FOUR
                  ONE
                THREE
               +THREE
               ______
               ELEVEN

John Gracefield


Date: 04/06/99 at 11:55:39
From: Doctor Rob
Subject: Re: Alphametric

Thanks for writing to Ask Dr. Math!

These problems can be reduced to a system of equations, by looking
at each column and considering carries:

(1) {E,F,H,L,N,O,R,T,U,V} = {0,1,2,3,4,5,6,7,8,9},
(2) None of F, O, T, or E = 0, because they are leading digits.
(3)     R + E + E + E = N + 10*a,
(4) a + U + N + E + E = E + 10*b,
(5) b + O + O + R + R = V + 10*c,
(6) c + F     + H + H = E + 10*d,
(7) d         + T + T = L + 10*e,
(8) e                 = E,
(9) 0 <= a,b,c <= 3, 0 <= d,e <= 2.

Now we have to use logic.  E can be 1 or 2, from (2), (8), and (9).

Case 1:  E = 2.

If E = 2, then e = 2 from (8), and from (7) it must be that T = 9,
L = 0, and d = 2.  Then

   (1a) {F,H,N,O,R,U,V} = {1,3,4,5,6,7,8},
   (3a) R + 6 = N + 10*a,
   (4a) a + U + N = 10*b - 2,
   (5a) b + 2*O + 2*R = V + 10*c,
   (6a) c + F + 2*H = 22,
   (9a) 0 <= a <= 1, 1 <= b <= 2, 0 <= c <= 3.

From (1a), (4a) and (9a), b = 1, so a + U + N = 8, and V is odd.
From (1a), (3a) and (4a), (R,N,a,U) = (1,7,0,1), (5,1,1,6), (7,3,1,4), 
or (8,4,1,3).  The first is impossible because U = R.  Thus a = 1.

Subcase 1.1:  R = 5, N = 1, U = 6.  Then

   (1b) {F,H,O,V} = {3,4,7,8},
   (5b) 2*O + 11 = V + 10*c,
   (6b) c + F + 2*H = 22,
   (9b) 0 <= c <= 3.

From (5b) and (1b), 1 <= c <= 2, V = 7, and (O,c) = (3,1) or (8,2).
Using that, (1b) and (6b), (F,H,c) = (4,8,2), so O = 8 = H, a
contradiction.  Thus R is not 5, N is not 1, and U is not 6.

Subcase 1.2:  R = 7, N = 3, U = 4.  Then

   (1c) {F,H,O,V} = {1,5,6,8},
   (5c) 2*O + 15 = V + 10*c,
   (6c) c + F + 2*H = 22,
   (9c) 0 <= c <= 3.

From (5c) and (1c), O = 8, V = 1, and c = 3, but then
F + 2*H <= 17 < 19, a contradiction. Thus R is not 7, N is not 3, and 
U is not 4.

Subcase 1.3:  R = 8, N = 4, and U = 3.  Then

   (1d) {F,H,O,V} = {1,5,6,7},
   (5d) 2*O + 17 = V + 10*c,
   (6d) c + F + 2*H = 22,
   (9d) 0 <= c <= 3.

From (5d) and (1d), (O,V,c) = (5,7,2) or (7,1,3), so 2 <= c <= 3.
Then F + 2*H <= 5 + 2*6 = 17 < 19 <= 22 - c, a contradiction.
Thus R is not 8, N is not 4, and U is not 3.

This means that Case 1 is impossible. Thus we are led to conclude
that E = 1, so e = 1. Then 0 <= a <= 1, b = 1, and we have

   (1e) {F,H,L,N,O,R,T,U,V} = {0,2,3,4,5,6,7,8,9},
   (2e) None of F, O, or T = 0, because they are leading digits.
   (3e) R + 3 = N + 10*a,
   (4e) a + U + N = 9,
   (5e) 1 + 2*O + 2*R = V + 10*c,
   (6e) c + F + 2*H = 1 + 10*d,
   (7e) d + 2*T = L + 10,
   (9e) 0 <= a <= 1, 0 <= c <= 3, 0 <= d <= 2.

From (5e), V is odd. From (1e), (3e), and (4e), we get (R,N,a,U) =
(0,3,0,6), (2,5,0,4), (4,7,0,2), (6,9,0,0), (7,0,1,8), or (9,2,1,6).

These give six subcases, which can be considered in turn, just like
those given above.

I leave the rest to you.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
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