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Finding the Digits of SEND + MORE = MONEY


Date: 05/19/98 at 23:34:59
From: Echo
Subject: A Basic Story Problem Of Guessing Mystery Numbers

Here is one of the more well-known cryptic problems, so I have been 
told, but I am stuck. I really don't know quite what to do or where 
to start. If this is one of the problems in your archives, please let 
me know. I looked, but was unable to find it.    

      SEND
    + MORE
    ------
     MONEY

I am trying to find what each of the letters represents. Thanks so 
much for your help.


Date: 05/20/98 at 13:04:05
From: Doctor Rob
Subject: Re: A Basic Story Problem Of Guessing Mystery Numbers

These problems can be reduced to systems of equations and 
inequalities:

   (1)  D, E, M, N, O, R, S, Y different whole numbers
   (2)  0 <= D <= 9
   (3)  0 <= E <= 9
   (4)  1 <= M <= 9    (it's the leading digit of "MORE", so not zero)
   (5)  0 <= N <= 9
   (6)  0 <= O <= 9
   (7)  0 <= R <= 9
   (8)  1 <= S <= 9    (it's the leading digit of "SEND", so not zero)
   (9)  0 <= Y <= 9
  (10)  D + E = Y + 10*a     (units column, carry a)
  (11)  N + R + a = E + 10*b (tens column, carry b)
  (12)  E + O + b = N + 10*c (hundreds column, carry c)
  (13)  S + M + c = O + 10*d (thousands column, carry d)
  (14)  d = M                (ten-thousands column)
  (15)  0 <= a <= 1
  (16)  0 <= b <= 1
  (17)  0 <= c <= 1
  (18)  0 <= d <= 1
  (19)  a, b, c, d are whole numbers

First of all, from (4), (14), and (18), 1 <= M = d <= 1, so:

  (18a) d = 1
  (4a)  M = 1

Now, in (13), S + 1 + c = O + 10.  If c = 0, by (8), (13), and (6),

   9 >= S = O + 9 >= 9

so S = 9 and O = 0. If c = 1, 9 >= S = O + 8 >= 8, so either S = 8 and 
O = 0, or S = 9 and O = 1. By (1) and (4a), O cannot be 1, so in 
either case:

  (6a)  O = 0
  (13a) S + c = 9

Looking at (12), (3), and (16):

  9 = 8 + 1 >= E + b = N + 10*c > 10*c

so c < 9/10, and by (17) and (19): 

  (17a) c = 0

and by (13a):

  (8a)  S = 9

From (12), (6a) and (17a), E + b = N, so, by (1):

  (16a) b = 1
  (12a) E + 1 = N

Now, from (11) and (16b), N + R + a = 10 + E. Putting this together 
with (12a):

  (11a) R + a = 9

which tells you by (1), (15), and (8a) that:

   (7a) R = 8
  (15a) a = 1

From (10) and (15a):

  (10a) D + E = 10 + Y

Now, consider the possibilities for Y. It must be 2, 3, 4, 5, 6, or 7.
Use (10a) and (12a) along with (1), (4a), (6a), (7a), and (8a).

So far we have:

   1 0 1 1
     9 E N D
  +  1 0 8 E
  ----------
   1 0 N E Y

From here, you can finish by yourself. For more assistance, refer to 
this answer from the Dr. Math archives:

   Send More Money
   http://mathforum.org/library/drmath/view/57951.html   

-Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
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