Send More Money

October 19th, 2005 admin Posted in Tidbits |

My father asked me this question over the phone this weekend.

“A college going kid’s father, tired of his sons repeated requests for money throws him a challenge. He asks the son to solve a math puzzle, before expecting any more money from the dad. SEND + MORE = MONEY is a mathematical equality, where each letter is a unique numeral. What is the equality in numbers?”.

It turned out to be an immensely interesting puzzle. It has a unique solutions, the solution can be logically deduced, and it makes a sensible English phrase! That third point is something I am still not sure whether to be amazed by or not.

Here is a sequence of steps I followed in my deduction of the solution

  S E N D
+ M O R E
---------
M O N E Y

The M has to be a 1, since the carry from adding S of SEND and M of MORE can be either a 0 or a 1, and in this case it is a 1 since we have M in MONEY to represent it. Another way to say the same thing is adding two numbers in the thousands can not give you twenty thousand or more. even 9999 + 9999 falls short of 20000 by 2.

  S E N D
+ 1 O R E
---------
1 O N E Y

S in SEND and 1 from 1ORE have to add to result in a carry since the whole existance of the 1 in 1ONEY depends on it. That is possible only if S is 9, there is no carry from the hundreds’ place (and therefore the O in MONEY is 0) or S is 8 and there is a carry of 1 from the hundreds’ place (in which case the O in MONEY is 0 again) or indeed the case that S is 9, and there is a carry of 1 from the hundreds’ place (and here the O in MONEY is 1). Since O can’t be 1, as M has wrested that privilege, the only two plausible choice is that S in SEND is 8 with a carry from the hundreds’ place or the S is 9 with no carry from the hundreds’ place. So, are we stuck? Which of these two to choose? Well, time to look a little into the future, by which I mean the hundreds place. One thing is certain, whether S is 9 with no carry or 8 with carry, the O in MONEY is 0. Therefore the O in MORE is 0 too! So looking at the hundreds’ place, E + 0 = N. That can’t be. There must be a carry from the tens’ place! So the hundreds’ place equation now reads, 1 + E + 0 = N. The question we are after is does 1+E give rise to a carry or not. Only way it can give rise to a carry is if E = 9. But then 1+E =10 and that means N = 0. That can’t be. O has gotten that privilege. So we do not know what E is but is certainly is not 9 (and not 0 or 1). So that means, since there is no carry over from the hundreds place S is 9.

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

N is one more than E. At first glance the values E can take are 2,3,4,5,6,7,8. Of these, 8 has to be ignored since then N would be 9. But 9 has already been assigned to another letter. Still the choices for E are too many. Looking at the tens’ place, N + R = E + 10 or 1 + N + R = E + 10 (if there is a carry from the ones’ place). The 10 in the above equation comes in because there is a carry from the tens’ to the hundres’ place for sure. Substituting E+1 for N, we get either E + 1 + R = E + 10, which implies R = 9 which is not possible, or that, 1 + E + 1 + R = E + 10, which leaves us with the information that R = 8 and equally importantly, that there is a carry from the ones’ to the tens’ place.

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

Now we look at the ones’ place. D, E and Y have to be three amongst 2,3,4,5,6,7 since the others have been taken. The information that the ones’ place lends a carry to the tens’ place means D + E = Y + 10. Note than the ones’ place cannot have a carry from the tenths’ place. This implies that D and E have to add up to at least 12, since Y is, at a minimum, 2. The only ways of adding a pair from the remaining numbers to get 12 or more are 5+7 or 6+7. E is either 5,6 or 7. Now we fall back upon the hundreds’ place to bail us out of the dilemma. The hundreds’ place tells us that E + 1 = N. Therefore E cannot be 7, since then N would be 8. N cannot be 8 since 8 is taken by R. Since E cannot be 7, E is either 5 or 6. Either way, D is 7. Now, if E were 6, N, which is E + 1, would be 7. That is not possible since we just allocated 7 to D. So E is 5 and N is 6.

  0 1 1
  9 5 6 7
+ 1 0 8 5
---------
1 0 6 5 Y

Y is therefore 2. And the final solution is 9567 + 1085 = 10652