# If each letter stands for a number

## 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:

## 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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UltrafastPED
Gold Member
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)

So, were my three initial observations correct or incorrect?

3) is definitely incorrect........

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.

UltrafastPED
Gold Member
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.

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?

UltrafastPED
Gold Member
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.

How is it clear that there is a carry over from the I+I column?

UltrafastPED
Gold Member
Otherwise R=E which violates the rules.

UltrafastPED
Gold Member
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.

I am rather confused. First, you write DE = R + 1. Secondly, what is the difference between N + D = R and
N + D = 1R?

Also, why isn't it R = E + 1, instead of DE = R + 1

UltrafastPED, I found out where you are wrong. E can't equal 10.

UltrafastPED
Gold Member
Well the rest is up to you!

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 $E \ne I \ne 0$, 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!

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.

I think I might have to ask my professor concerning this problem.

What is the maximum carry over when two numbers less than 10 are added?
From this you can get both R, D and E.

OmCheeto
Gold Member
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:

The O.P. needs to figure something out himself...

OmCheeto
Gold Member

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.........

:grumpy:

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

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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?

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.

ehild
Homework Helper
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

UltrafastPED
Gold Member
Or just see message #8 from yesterday!

What precisely do you mean by this? Have I not been able to determine so portions of the problem myself?
Woops, sorry, no offence . I'm really pissed at this thread as I already gave a quite detailed explanation with the answer and a mentor spotted it, deleted it and sent me a 'notice' for not complying with PF rules.:grumpy:

Woops, sorry, no offence . I'm really pissed at this thread as I already gave a quite detailed explanation with the answer and a mentor spotted it, deleted it and sent me a 'notice' for not complying with PF rules.:grumpy:

No, need to worry, I wasn't offended; I just wanted to make it clear that I was, to some degree, trying the answer myself. Really, they deleted the post? That's unfortunate.

I want to thank y'all earnestly for your aid.