### Video Transcript

Each letter in this cryptarithmetic puzzle represents a different digit, and none of the numbers use leading zeros. Find the value of each letter in the puzzle in order to make a correct sum.

We’re gonna solve this puzzle using logic and elimination. First, I’m just gonna move this over to make some space. And then we’re going to list out each of the letters C, R, O, S, A, D, N, G, E. We list out all of the digits that each of these letters could represent.

Almost immediately, we can look at this D and determine what number it must represent. To find the value of D, we’ll need to add C and R. But even if C and R were the largest of the digits, for example, nine and eight, they would add up to 17. So no matter what the digit D represents will have to be equal to one. It cannot be greater than one.

We also know there are no leading zeros, which means that D cannot be equal to zero. D must be equal to one. And that means D is not equal to any of these other numbers. It also means that none of the other letters can be equal to one.

That’s a great start. But unfortunately, that’s really the only one that’s immediately obvious to us. But I do recognize that S plus S equals R. And we can draw some conclusions from this. If we know that something plus itself equals R, we can eliminate some of the values. Because R is one digit, S must be less than five. Five plus five would be equal to 10. And 10 has more than one digit.

We don’t know what S is, but we do know some things that S could not be. It can’t be larger than five. But what about zero? Could S be equal to zero? No, that’s not possible. Zero plus zero would equal zero. Then the statement would have to say S plus S equals S. We have three options for our S value. And because we have three options for our S value, that also means we have three options for our R value.

Our first option, two plus two equals four. That means R could be equal to four. Three plus three equals six. R could be equal to six. And four plus four equals eight. These are the only three options for R. And we can eliminate all the other digits. Again, we don’t know what R is, but we do know some things that R isn’t.

Moving on to S plus D equals E, S plus D equals E and we already know that D equals one. Here again, we don’t know what S is, but we do know its three options. One option is two plus one equals E. E could be equal to three, or three plus one equals E. In that case, E would be four. The final option for E is four plus one. E could be equal to five. E cannot be equal to zero, two, six, seven, eight, or nine.

We said a few things about C plus R in the beginning, but we’ll look closer at C plus R now. C plus R equals something with a one in the tens place. If we look to R, we’ve eliminated all but three possible options. R would be equal to four, six, or eight. If R equals four, what could C be equal to? Let’s try six. C equals six. Six plus four equals 10.

Okay, in this scenario, C equals six and A equals zero. But if A equals zero, then we have a problem, because anything plus zero equals itself. What would happen would be the letter O plus zero would have to be equal to the letter O and not the letter G. What does this tell us?

First of all, it tells us that A cannot be equal to zero. It also tells us that C cannot be equal to six. Okay, but that’s not the only case when R is four, is it? How about seven? Could C be equal to seven?

Well, seven plus four equals 11. And because we know that D represents the one, seven plus four would equal DD, which means C cannot be equal to seven. Is there anything else we can add to four? We could try eight. Eight plus four equals 12. And so far, we have no reason to eliminate these two as options. So we can circle C is maybe equal to eight and A is maybe equal to two.

We have one more case if R equals four. That’s the case where C equals nine. Nine plus four equals 13, and we’ll circle nine for C as an option and three for A as an option. So what we’ve just said is that if R equals four, C could be eight or nine and A could be two or three.

Now we want to look at the cases where R equals six. Starting with four, four plus six equals 10. But we’ve already seen that A cannot be equal to zero. And that means C could not be equal to four. This is not a valid option.

We could try five. Five plus six equals 11. We’ve already seen that A cannot be equal to one. And if A can’t be equal to one, C cannot be equal to five. We’ve already eliminated C equals seven. C equals eight still seems like it could work. Eight plus six equals 14. If C equals eight and R equals six, then A would be equal to four. Those are all the cases for R equals six.

Now we want to examine the case if R equals eight. If C equals two, two plus eight equals 10. We know that A is not zero. Therefore, C is not equal to two. Three plus eight equals 11, and A cannot be equal to one. Therefore, C cannot be equal to three. We’ve already eliminated four, five, six, and seven as choices for C.

Our next option would be eight. Eight plus eight equals 16, but did you spot the issue here? If R equals eight, then C could not be equal to eight, which means that’s not an option for us. The last case would be C equal to nine. Nine plus eight equals 17. If C equals nine, R equals eight, then A would be equal to seven. And now what?

If we look closely, we’ll see that we’ve actually missed one of our pairs. We’ve seen the case where C equals eight and R equals six. What about the case where C equals nine? If C equals nine, then A would be equal to 15. And so now we have a fifth option for what A could be.

At this point, it can feel like we’re getting more and more options instead of less and less. But we go ahead and cross out what we know A cannot be. On my list, the letter C, we see that we can have zero, eight, or nine.

Let’s consider the case if C was equal to zero. If C was equal to zero and R can only be four, six, or eight, we would never end up with a one as the D. And that helps us eliminate the zero as an option for C.

At this point, I want to take all of the small addition problems and write them down horizontally. And then I’m going to try all the cases where C equals eight and C equals nine. Here we go.

C equals eight and R equals four. There we have 12. If R equals four, we put the four again. We don’t know O; we don’t know N; we don’t know O; but A would be equal to two, and we still wouldn’t know G. In this case, S has to be equal to two. Two plus one equals three. E would be equal to three. And two plus two equals four. This set didn’t help us eliminate anything, so we’ll just leave it there. And then we’ll move on for C being nine and R being four.

Nine plus four equals 13. Four plus O equals N; O plus A equals G, but we know that A equals three. In this case, S is equal to two and D equals one. Two plus one equals three. But look at what’s happened. In this case, both A and E are showing up as three. And we know that there cannot be duplicate digits. A cannot be equal to three.

This time, we’ll choose C equals eight and R equals six. Eight plus six equals 14. Six plus O equals N. We can’t solve that. O plus four equals G. S plus D, three plus one, equals four. And we’ve run into an issue because now we have A and E showing up as four. We can eliminate A equal to four.

The next round, C equals nine and R equals six. Nine plus six equals 15. Six plus O equals N. We can’t say anything about that. O plus five equals G. Three plus one equals four, and three plus three equals six.

What I wanna do now is write out which digits we’ve used in both of these cases. As I’m writing all of this down, I notice, and maybe you have too, a mistake in this set. Look at how many times the two is used. That’s because we’ve used the value two for both A and S in this instance. A cannot be equal to two, and this option is not valid.

Now that we’re down to this pink set, I want to write out all the digits we’ve already used. We’ve used the values one, three, four, five, six, and nine, which means we still need the values two, seven, and eight. We have three missing letters: O, E, and G. Is there a way to use two, seven, and eight and make all of these statements true?

If I plug in two for O, then I’ll end up with six plus two equals E. The two seems to work. Six plus two equals eight. And the last question, “Is two plus five equal to seven?” It is. It now seems like we’ve filled all of these spaces. Let’s go through and make sure we haven’t made any errors.

We’ve said that C equals nine, and that means no other value can be equal to nine. We’ve made R equal to six. So far so good. We’ve already identified D equals one. Now we have A equals five. Next, O equals two, G equals seven, N equals eight, S equals three, and E equals four. C equals nine; R equals six; O equals two; S equals three; A equals five; D equals one; N equals eight; G equals seven; E equals four.

If you’ve made it to the end, you’re probably wondering, “Is this the only way to solve this? Is this the best way to solve this?” This is the way that you solve a problem like this by hand. Today, most often people use computer programs to lessen the time it takes to solve something like this. Computers are able to do the process we just completed, but faster and more efficiently. However, if you’re ever stranded on a desert island and you need to solve a cryptarithmetic puzzle, this would be the way to do it.