# Video: Solving Cryptarithmetic Puzzles

Each letter in this cryptarithmetic puzzle represents adifferent digit, and none of the numbers use leading zeros. After finding the value of the digit that each letter represents inorder to make this a correct sum, work out which number would be represented by the word STELLAR.

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### Video Transcript

Each letter in this cryptarithmetic puzzle represents a different digit, and none of the numbers use leading zeros. After finding the value of the digit that each letter represents in order to make this a correct sum, work out which number would be represented by the word STELLAR.

We can remind ourselves of the rules of these type of puzzles. First, each letter represents a digit from zero to nine. And in the puzzle, that letter will always represent the same digit. And in a proper cryptarithmetic puzzle, there’s only one unique solution. To solve this, we’re going to use number sense, logic, and a bit of trial and error.

A good first step is listing out all the letters and then make a grid with all the digits each letter could be. We know that there are no leading zeros, which means the S, U, and P here cannot be zero. On our grid, we can eliminate zero from S, U, and P.

If we think carefully, there’s something else we can say about the digit that P represents. If we let S be nine and U be eight and if we say that one is carried over from the previous column, we have the maximum amount that we would ever have in this position. It would give us 17 without the carried one or 18 with the carried one. But that wouldn’t work because L has to be a different digit than U; they can’t both be eight.

What this is telling us is that the amount that P must be is a one. Using these digits, we could never make this two or more. And so, we’ll say that P is one. On our grid, that means that P can be nothing else. And it means no other letters can represent one. When I’m solving the puzzles, I like to create a second set of boxes that I can use for filling in and trial and error so that I don’t have to mark up the original.

The only thing we can fill in so far is that P is one. We can now move our attention to the digits place where we have N plus S equals S. Remember that N and S have to be digits between zero and nine. If N plus S equals S, we can say here that N must be equal to zero. Zero is the only thing you could add to S and have S as a result. We’ll mark N equals zero. Nothing else can be zero now. And then, in our boxes, we’ll fill in zero everywhere the N occurs. That’s three places.

And then, we notice that we have a really interesting situation here. It almost looks the same as the first column. We have A plus R equals A. But we know that R can’t equal zero because N already equals zero. What then could make A plus R equal to A. Well, the column previously has a result of zero. But we’re not adding two values that equal zero. And what that tells us is that T plus A, the digits in this column, must be equal to 10.

The zero goes here. And a one will be carried over. We can write this situation out like this. One plus A plus R has to equal A plus 10. What we’re saying is there has to be another thing carried over. We know this because we’re adding R and one to A. And therefore, the result could not just be the digit A. It has to be more than that.

If we look at this, we see that there’s an A on either side of this equation. And we can cancel that out. And that says one plus R equals 10. If we subtract one from both sides, we found that R must be nine. There are two Rs in our puzzle. So, we fill in the nines where they go and complete our grid. Now, that R doesn’t necessarily lead us to any other letters. So, we go back to our puzzle and see if we can recognize any other patterns.

Now, we know that N equals zero. But U plus N equals E. If U plus N does not equal U, it tells us that we need one carried over from the previous side. And it also tells us that E is going to equal U plus one. And looking at R plus U equals T. That tells us that nine, which is R, plus U has to equal T plus 10, since there’s a carryover. By subtracting nine from both sides, we see that U equals T plus one. Using this, we can say we have some value for T, and then we have T plus one which is U. And we have U plus one, which we could write as T plus one plus one, T plus two, which is E.

So, we now have T, U, and E in order, T being the smallest and E being the largest. Since E has to be two digits bigger than T and currently the smallest thing T could be is two, we can eliminate two and three from E’s range. And since E is two bigger than T, that means that T cannot be eight or seven. If T was eight, then E would have to be 10. And that’s not possible. If T was seven, then E would have to be nine. And nine is already taken.

We’ll do the same thing for U. Since U is one larger than T, U cannot be two. And since the largest thing T could be is six, U cannot be eight since it is one more than T. At this point, there doesn’t seem to be much else we can do by elimination. So, what we want to do is look at the letters who have the least options. So, there are five options for T, five options for U, and five options for E. We want to choose one of those letters and begin testing.

Since U occurs first in the puzzle, let’s try U equals three. If we plug in three for U, we want to put that three everywhere a U occurs. Nine plus three is 12. We carry our one. One plus three plus zero equals four. Under these conditions, we have a two where the T is. So far, there’s no problems. If we look at this box, we know that T plus A has to equal 10. And if two equals T, then A would be eight. And there are two more As in the next column. We now have two plus eight equals 10, carry our one. And then, we have one plus eight plus nine equals 18.

Because we have a one here, we know that whatever goes here needs to equal more than 10. We have one plus three. We could plug in six. Then, we have one plus six plus three. But that equals 10. And this is not a zero. So, okay, let’s try a seven. One plus seven plus three equals 11. But we don’t have another P in this digit. This is an L. And we can’t try eight because eight is not S; eight is A. And we’ve already used nine. And so, we’ll go back to the U and say U cannot equal three. And if U cannot equal three, T cannot equal two and E cannot equal four.

Now, we’ll clear out everything we don’t know for certain, we’ll go back to our U, and choose another value. The next highest value for U is four. We plug in four everywhere for U. Nine plus four is 13. One plus four plus zero is five. T equals three. And we know that three plus some value has to equal 10. So, we plug in seven. If A equals seven, then we have a seven here and here. So far, again, no problems. We need one plus four to be greater than 12. We could plug in eight. Eight plus one plus four equals 13. And again, we have a problem in this position because L and T cannot both have the digit three. So, U can’t be four, T can’t be three, and E cannot be five. And we clear the deck to start again.

I wanna stick with the letter U for trial and error because that’s consistent. But because we’ve been having problems with the lower values for U, this time, I’m going to try U as seven. Plug in seven everywhere we see the U. Nine plus seven equals 16. One plus seven plus zero equals eight. And we found that T equal six. We know that six plus something has to equal 10. That makes A four. And if A is four, we have two fours in this column. One plus four plus nine equals 14. And we’re back to where we keep running into problems.

We need one plus some value plus seven to be greater than or equal to 12. We know that this is true because they can’t equal 10 and they can’t equal 11 as those digits, the zero and one, have already been used. I know that one plus four plus seven equals 12, but we’ve already used the four in this working. So, we could try five. One plus five plus seven equals 13. And so far, we have not repeated a digit. And what we’ve done is we found S equals five, which makes the two blanks in our first column five. Zero plus five equals five.

Here, we think we have a correct solution. For our first tech, let’s take what we think is SATURN and what we think is URANUS and add them together. Zero plus five is five. Nine plus seven is 16, carry your one. One plus seven plus zero is eight. Six plus four is 10, carry your one. One plus four plus nine is 14, carry your one. One plus five plus seven is 13.

And then, we want to check the letters. We have S equals five. And that is located in the three places that the Ss are found. A equals four. And that is located in the three places the A are found. T equals six. And there are two sixes and two Ts in the puzzle. U equals seven. And there are three sevens and three Us in the correct position. R equals nine. N equals zero. And then, P equals one. L equals three. E equals eight.

From there, I like to write the solutions next to the letters, but our work here is not quite finished since we want to know what number the word STELLAR would be. S is five. T is six. E is eight. L is three. So, we have three three. A is four. And R is nine. The word STELLAR is then 5,683,349. And you’ve made it to the end of the puzzle. Now, you might be ready to try one of your own using logic, number sense, and a bit of perseverance.