Cracking a cryptarithm by reasoning about digits, not guessing

0:00A calculation with letters standing where the digits should be. Every letter is one digit, and the job is to work out which. There are three rules, and they are not decoration. One. A letter is the same digit everywhere it appears. Two. Two different letters can never be the same digit. Three. No number in the puzzle starts with a zero. That second rule is worth reading slowly. It forbids two letters from sharing a digit. It does not say every digit needs a letter, which is why bare digits turn up inside these puzzles quite legally.

0:41And the whole thing looks like a search through thousands of possibilities. It is not. Place value is what makes it solvable. Write the calculation in columns, the way you would work it by hand. Now each column is almost its own small problem. The units column involves only the units digits. The tens column involves only the tens digits, plus whatever the units column handed upwards. That is the whole structure. A column touches its own digits and one neighbour, and nothing else at all.

1:21So this is not a haystack. It is a chain of local facts. Work the chain, and the answer falls out of it. Before any puzzle, one thing needs bounding. How much can a column pass to the next one? Add two digits. The most you can have is nine and nine and a carry, which is nineteen. So the column passes at most one. Add three digits. Nine and nine and nine, plus a carry of two, is twenty-nine. So a column passes at most two.

1:56Multiply, and it gets bigger. A digit times eight can reach seventy-two, so that column can pass as much as seven. The carry is one is a habit, not a fact. Bound it from the operation in front of you. Now a puzzle. A two-digit number written A then one, added to a two-digit number written one then B, giving a two-digit answer written B then zero. Start at the units column. Always start there.

2:29One plus B has to end in zero. So B is nine, and that column passes a one upwards. Now the tens column. A, plus one, plus the one carried up, has to come out as B, which we already know is nine. So A is seven. Seventy-one plus nineteen is ninety. Check it. A is seven, B is nine, no two letters share a digit, and nothing starts with a zero.

2:58Two columns, and not one guess anywhere. Here is one where the same number appears three times over. A two-digit number, O then N, written down three times and added, giving a two-digit answer, P then O. The units column is N plus N plus N, and whatever it comes to has to end in O. Work it through, and there are three answers here, not one. Seventeen tripled is fifty-one. Twenty-four tripled is seventy-two. Thirty-one tripled is ninety-three.

3:34All three obey every rule completely. Which is worth knowing, because finding one answer and stopping is answering a smaller question than the one you were asked. Multiplication brings a new kind of constraint, and it is a strong one. A two-digit number, P then Q, multiplied by eight, giving a two-digit answer, R then S. Before touching a single digit, look at the sizes. A two-digit number multiplied by eight is usually three digits. Thirteen eights are a hundred and four.

4:13So there are only three two-digit numbers that stay two digits. Ten, eleven and twelve. That is the entire search, and counting digits is what got us there. Now the rules finish it off. Ten eights are eighty. The units digit of the number is zero and so is the units digit of the answer, so Q and S would be the same digit. Forbidden. Eleven eights are eighty-eight. But then P and Q would both be one. Forbidden.

4:47Twelve eights are ninety-six. P is one, Q is two, R is nine, S is six. All different, and nothing starts with a zero. One answer, and every step that got there was a consequence. Notice how much of the work the size bound did. Three candidates instead of ninety. Then two rule violations, and it was finished. The other sharp tool is the units digit of a product. Whatever the other digits are doing, the units digit of the answer is decided entirely by the units digit you started with.

5:27Multiply anything at all by six, and look at what lands in the units place. Zero, two, four, six, eight. Every one of them even. Six times anything is even, so the units digit of the answer is even, always, with no exceptions. That single observation kills half the digits before you begin. But it does not work for every multiplier. Multiply by three and the units digits come out both even and odd, so you have learnt nothing.

6:03Know which of your tools is actually cutting. Now a harder one. A two-digit number, G then H, multiplied by H, the same letter as its own units digit, giving a two-digit answer, nine then K. So the answer is somewhere in the nineties. Seven products land in the nineties. Eleven nines are ninety-nine, twelve eights are ninety-six, forty-six twos are ninety-two, twenty-four fours are ninety-six, forty-seven twos are ninety-four, thirty-one threes are ninety-three, and sixteen sixes are ninety-six.

6:45First condition. The multiplier has to be the units digit of the number itself. That leaves two of them. Twenty-four fours, and sixteen sixes. Second condition. No two letters share a digit. Sixteen sixes makes H a six, and K a six as well. So it is twenty-four fours are ninety-six. G is two, H is four, K is six. Seven down to two down to one, and every step was a rule rather than a hunch.

7:19One more, with three letters this time. A three-digit number, B, Y, E, multiplied by six, giving a three-digit answer, R, A, Y. Count the digits first. If B were two, then two hundred sixes are already twelve hundred, which is four digits. So B is one. That is settled before anything else is known at all. Next, Y. A hundred and seventy sixes are a thousand and twenty, four digits again. So Y is at most six.

7:53And Y is the units digit of the answer, which is six times E, which we just said is always even. Even, and at most six. That leaves four candidates, and only one survives. A hundred and five sixes are six hundred and thirty. It is worth seeing how much those rules are actually carrying. Take all fifteen puzzles here, and switch off the rule that two letters cannot share a digit.

8:24Five of them stop having a single answer. Switch off the rule against leading zeros instead, and three of them break. The eight puzzle on its own goes from one answer to eight. They are not politeness. They are conditions, and without them most of these are not puzzles at all. The first rule, that a letter is the same digit everywhere, is not optional in the same way. It is what makes a repeated letter the most useful thing on the page.

8:57Two Ps, three Gs, a letter that turns up twice. That is where to look first, every time. Last thing. When you have an answer, check it properly. Does the arithmetic come out? Is every letter one digit throughout? Are all the letters different from each other? Does anything start with a zero? All four, every single time. And then the question people skip. Is it the only one? Of the fifteen puzzles here, twelve have exactly one answer and three have more than one.

9:34The tripled one had three. Another has two. If you stop at the first, you have answered a different question. So the method is this. Bound the sizes. Read the units column. Use the rules to eliminate. Then ask whether anything else survives. There is no guessing anywhere on that list.