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If each letter stands for a number

  1. Sep 7, 2013 #1
    1. The problem statement, all variables and given/known data
    If each letter stands for a number from 0 to 9 (and different letters stand for different
    numbers), find the numerical value of each letter in this addition:


    2. Relevant equations



    3. The attempt at a solution
    (1) N + D = R

    (2) I + I = E ----> I = E/2, which means E must be even

    (3) R = E

    These were my initial conjectures. However, I now feel that they are wrong. At first, I thought that there was no carrying over of digits--which would be a helpful fact--; then I realized that D was in the 1000's place, which implies that there must have been some carrying over.

    Could someone help me?
     

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    Last edited: Sep 7, 2013
  2. jcsd
  3. Sep 7, 2013 #2

    UltrafastPED

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    Since the problem is in base 10 (digits are 0-9) we can avoid carrying by expanding the numbers thus:
    (1000*D +100*E + 10*E + R) = (10*I + N) + (100*R + 10*I + D)
     
  4. Sep 7, 2013 #3
    So, were my three initial observations correct or incorrect?
     
  5. Sep 7, 2013 #4
    3) is definitely incorrect........
    Start with D. :wink:
     
  6. Sep 7, 2013 #5
    What do you mean by "start with D?" Yeah, I figured that 3) would be wrong, seeing as they don't meet the requirement of being distinct numbers.
     
  7. Sep 7, 2013 #6

    UltrafastPED

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    The problem is a puzzle - it resolves to 999D + 110E = 99R + 20I + N where D,E,R,I,N are five distinct digits.

    At this point you can try different combinations for a while ... a small computer program would find any possible solutions very quickly. Of course there may be something very clever that can be done, but I prefer algorithms to cleverness - they get the work done.
     
  8. Sep 7, 2013 #7
    Now, can I take the equation 999D + 110E = 99R + 20I + N, solve for N, and choose values for D, E, R and I such that N is an integer and N does not equal one of the other letters?
     
  9. Sep 7, 2013 #8

    UltrafastPED

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    Yes ... though you actually need to provide a guess for all five at one time since there is only one equation!

    If you are clever you may find a way to generate more equations ... but I don't; that's why I suggested a small computer program to cycle through all of the possible combinations. In other words, guess and check!

    Looking at the original expressions it is clear that there is a carry from the I+I column - therefor D=R+1. That gives you another equation to work with. And to get a carry from 2I (an even number) you must have I>4 (because a carry from the previous column can add no more than 1). We also know that D is the result of a carry, hence D=1.

    So I guess that cleverness can simplify things after all - all I needed was a bite to eat and I began to see things.

    Now you can work through the rest ... and when you cannot go any further, guess the rest.
     
  10. Sep 7, 2013 #9
    How is it clear that there is a carry over from the I+I column?
     
  11. Sep 7, 2013 #10

    UltrafastPED

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    Otherwise R=E which violates the rules.
     
  12. Sep 7, 2013 #11

    UltrafastPED

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    Since D is the result of a carry, D=1; but DE = R + 1 (from the carry from I + I) ... thus R=9, D=1, and E=0.

    Thus DEER = 1009.

    I doubt that my method is good enough for the Putnam exam - but I do keep plugging away! I hope that you are having as much fun as I am. I now notice that the first column has N+D=R or maybe N+D=1R if there is a carry ... but D=1, R=9 ... so N=8.

    That means there is no carry so I + I = 1E =10, so I = 5.

    So 58 + 951 = 1009. So it can be solved by paying attention - thus proving that computers are not always required.
     
  13. Sep 7, 2013 #12
    I am rather confused. First, you write DE = R + 1. Secondly, what is the difference between N + D = R and
    N + D = 1R?
     
  14. Sep 7, 2013 #13
    Also, why isn't it R = E + 1, instead of DE = R + 1
     
  15. Sep 7, 2013 #14
    UltrafastPED, I found out where you are wrong. E can't equal 10.
     
  16. Sep 7, 2013 #15

    UltrafastPED

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    Well the rest is up to you!
     
  17. Sep 7, 2013 #16
    All right. Let's recap. D must be one, because it is the result of a carry over, so no other letter can assume the value 1. E must be even, because 2I = E; however, we know that [itex]E \ne I \ne 0[/itex], otherwise I = E, which doesn't meet the conditions specified in the problem. If we solve for N, then we get the function N(E,R,I) = 999 + 110E -99R -20I. Before I proceed with substituting in random values, how many combinations do we have? I know there are 4 choices for E. Does that leave 8 choices for R, and 7 choices for I?

    Oh wait!!!!!!! E can't exceeed 10, meaning we can't choose any numbers exceeding 5 for I. This means that I can be 2, 3, or 4.

    Let me work out the rest of the details. Feel free to respond before I do!
     
  18. Sep 7, 2013 #17
    Hmmm, this doesn't seem to be working. If I = 2, then E = 4. So, are function is N(4,R,I). Let R = 9, then

    N(4,9,4) = 508, which isn't an integer between 0 and 9.
     
  19. Sep 7, 2013 #18
    I think I might have to ask my professor concerning this problem.
     
  20. Sep 7, 2013 #19
    What is the maximum carry over when two numbers less than 10 are added?
    From this you can get both R, D and E.
     
  21. Sep 7, 2013 #20

    OmCheeto

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    You forgot I.

    :tongue:
     
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