If each letter stands for a number

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  • #1
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Homework Statement


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:


Homework Equations





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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Answers and Replies

  • #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)
 
  • #3
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So, were my three initial observations correct or incorrect?
 
  • #4
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3) is definitely incorrect........
Start with D. :wink:
 
  • #5
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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.
 
  • #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.
 
  • #7
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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?
 
  • #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.
 
  • #9
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How is it clear that there is a carry over from the I+I column?
 
  • #10
UltrafastPED
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Otherwise R=E which violates the rules.
 
  • #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.
 
  • #12
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I am rather confused. First, you write DE = R + 1. Secondly, what is the difference between N + D = R and
N + D = 1R?
 
  • #13
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Also, why isn't it R = E + 1, instead of DE = R + 1
 
  • #14
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UltrafastPED, I found out where you are wrong. E can't equal 10.
 
  • #15
UltrafastPED
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Well the rest is up to you!
 
  • #16
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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!
 
  • #17
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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.
 
  • #18
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I think I might have to ask my professor concerning this problem.
 
  • #19
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What is the maximum carry over when two numbers less than 10 are added?
From this you can get both R, D and E.
 
  • #20
OmCheeto
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What is the maximum carry over when two numbers less than 10 are added?
From this you can get both R, D and E.
You forgot I.

:tongue:
 
  • #21
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:biggrin:
The O.P. needs to figure something out himself...
 
  • #22
OmCheeto
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:biggrin:
The O.P. needs to figure something out himself...
From my notes from this morning, I was only able to initially logically deduce the values of R & D in my brain. Your brain must be bigger than mine. hmmm.......

And why is this post in "Calculus & Beyond Homework" section? This belongs in either the "Fun, Photos & Games" or "Precalculus Mathematics Homework" section. I'd push that gosh darned report button, but it never says anything about misplaced questions.......

oh poop. They changed the wording.....

... or for other reasons.
I didn't get the memo.........

Admins*..........

:grumpy:

--------------------
* always changing the rules, and never informing the flotsam........

I should probably PM lisab about this thread. She might enjoy a 30 second mathematical recreation.
 
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  • #23
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:biggrin:
The O.P. needs to figure something out himself...
What precisely do you mean by this? Have I not been able to determine so portions of the problem myself?
 
  • #24
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What is the maximum carry over when two numbers less than 10 are added?
From this you can get both R, D and E.
The maximum carry over is one, which is a fact I used in my solutions.
 
  • #25
ehild
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You know already that D=1, R=9 and E=0. Substitute the letters with these numbers in the sum.


......I N
...9 I 1
--------
1 0 0 9


The sum of two one-digit numbers can not exceed 9+9=18. Therefore N+1 = 9. N?
As there is no carry at N+1, I+I=0 or I+I=10. But I cannot be 0. I+I carries 1. So what is I?

ehild
 

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