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Date: 04/18/2002 at 21:11:51
From: George Hollinger
Subject: Changing a letter problem into a math equation


I have a problem that I cannot fiqure out. I have tried logical 
reasoning and can't get it.

  + MORE

I have to change the letters into numbers, and all the same letters 
have to be the same number. I hope you can help!

George, age 11

Date: 04/19/2002 at 02:59:50
From: Doctor Jeremiah
Subject: Re: Changing a letter problem into a math equation

Hi George.

The secret is to notice that the answer has more letters (5 letters) 
than the question (4 letters).

That M at the beginning of money is a carry from the thousands place, 
so M = 1.  Now we have:

   + 1ORE
  = 1ONEY

Now, in the thousands place there is a 1, so the only value for S that 
could cause a carry is S = 9 and that means O = 10.  Now we have:

   + 10RE
  = 10NEY

Now look at the hundreds place. If there were no carry from the tens 
place, E and N would be the same because E+0 = N, but E and N can't be 
the same, so there must be a carry from the tens place.  Now we have:

    1 1    <-- carry
   + 10RE
  = 10NEY

Now the equation for the hundreds place is 1+E+0 = N or just 1+E = N. 
In the tens place we can have N+R = E+10 if there is no carry from the 
ones place, or we can have 1+N+R = E+10 if there is.

First test: no carry from the ones place:

N+R = E+10  and  1+E = N
(1+E)+R = E+10
1+R = 10
R = 10-1
R = 9

But S = 9, so R cannot = 9.  That means there is a carry from the ones 
place and we get:

1+N+R = E+10  and  1+E = N
1+(1+E)+R = E+10
2+R = 10
R = 10-2
R = 8

So now we have:

    1 11   <-- carry
   + 108E
  = 10NEY

N cannot be 0 or 1 because 0 and 1 are taken.
N cannot be 2 because 1+2+8 = 11 and then E would equal 1, but it 
cannot equal 1 because 1 is taken.
N could equal 3,4,5,6 or 7 but it cannot equal 8 or 9 because 8 and 9 
are taken (and E must be 2,3,4,5 or 6 because it is 1 smaller than N).

If E were 2, then for the ones place to carry D would have to be 8 or 
9, and both 8 and 9 are taken, so E cannot be 2 (and N cannot be 3).

If E were 3, then for the ones place to carry D would have to be 7,8, 
or 9, but D cannot be 7 because then Y would be 0, which is taken, so
E cannot be 3 (and N cannot be 4).

If E were 4, then for the ones place to carry D would have to be 
6,7,8, or 9.  D cannot be 6 because Y would be 0, but D cannot be 7 
because Y would be 1, and 0 and 1 are both taken, so E cannot be 4 
(and N cannot be 5).

The only two possibilities for E now are 5 and 6.

If E were 6, then N would be 7 and D would have to be 4 (which would 
make Y = 0), 5 (which would make Y = 1), 6 (which is taken by E), or 
7 (which is taken by N). There are no solutions for E = 6, so E must 
be 5.

So now we have:

    1 11   <-- carry
   + 1085
  = 1065Y

Do the same reasoning for D and Y and get the answer.

If you need more help check out this archived answer:

   Finding the Digits of SEND + MORE = MONEY

- Doctor Jeremiah, The Math Forum
Associated Topics:
High School Puzzles
Middle School Puzzles

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