Arranging given digits to make the largest product

0:00Here is a frame with three empty boxes. Two of them make a two-digit number, and the third one stands outside, multiplying it. You are given three digits: two, three and five. Each one goes in exactly one box, and you want the product to come out as large as you can make it. There is a temptation here, and almost everybody feels it. Just try them all and see which wins.

0:28You could. There are only six ways. But then you would know the answer and not know a single thing about why. So this time nothing gets multiplied out until the very end. Every step is a comparison, and the comparisons decide it. First, how do we know there are six? Not by writing them out and counting, because a list you wrote by hand is a list you might have missed something from.

0:58Build it instead. The first box can take any of the three digits: that is three choices. Whichever you put there, two digits remain, and they can go into the two boxes that are left in either order. So each of those three choices splits into two. Three, each splitting into two, is six. And now six is not a number you were told; it is a number you watched being built.

1:26Here they are, all six, and every one uses each digit once. Now sort them, not by size, but by which digit is doing the multiplying. Two arrangements have the two outside. Two have the three outside. Two have the five outside. That is not a tidy-up. It is the whole first move, because inside a group the multiplier is the same in both. And when two things are being multiplied by the same number, the bigger one wins. Nothing about digits, nothing about place value. Just that.

2:07Take the group with two outside. Thirty-five times two, and fifty-three times two. Fifty-three is larger than thirty-five, and both are being doubled, so fifty-three times two is the larger product. We have not worked out either of them. The same cut lands in the other two groups. Fifty-two beats twenty-five. Thirty-two beats twenty-three. Six arrangements, three comparisons, and three of them are gone. Three are left standing: fifty-three times two, fifty-two times three, and thirty-two times five.

2:47One more falls, and it falls to the same idea used the other way round. Fifty-three times two, against fifty-two times three. The two-digit parts are almost the same size. Fifty-three, fifty-two. But one is being doubled and the other is being tripled. The tiny advantage in the two-digit number cannot make up for a whole extra copy of it. So fifty-three times two goes. Two arrangements are left, and here the easy comparisons run out.

3:21Fifty-two times three, and thirty-two times five. The larger two-digit number has the smaller multiplier. Neither one is bigger in both ways, and instinct is finished. So take a product apart instead. Thirty-two times five. The thirty-two is three tens and two ones, so the product is five lots of three tens, plus five lots of two. Five lots of three tens is one hundred and fifty. Five lots of two is ten. Together, one hundred and sixty.

3:59Look at those two pieces beside each other. One hundred and fifty, and ten. They are not remotely the same size. That is not an accident of this example. One piece is always ten times the other's scale, because one of them counts tens and the other counts ones. And the big piece has a shape worth noticing. It is ten, times the multiplier, times the tens digit. Ten times a product of two of your digits.

4:33Now split the other one the same way. Fifty-two times three is three lots of five tens, plus three lots of two. That is one hundred and fifty, and six. One hundred and fifty again. The big pieces are identical. And you can see why, once you look at which digits landed where. In thirty-two times five, the big piece is ten times three times five. In fifty-two times three, it is ten times five times three.

5:06The same two digits, three and five, in the two slots that carry the weight, just swapped between them. Swap two numbers being multiplied and the answer does not move. That is the only reason those pieces agree, and it is worth saying out loud, because everything now depends on it. If the big pieces are equal, they cannot decide anything. Whatever separates these two, it is in the small pieces.

5:38Ten, against six. So thirty-two times five wins, by four. One hundred and sixty against one hundred and fifty-six. And now look at what the small piece is made of. It is the leftover digit, the two, multiplied by whichever digit went outside. The two is stuck. It was always going to be the one left over. The only choice is what it gets multiplied by. Multiply it by five and you get ten. Multiply it by three and you get six. So the largest digit goes outside.

6:18None of that used the fact that the digits were two, three and five. Call them a, b and c, smallest to largest, and run it again. Group by the multiplier: the larger two-digit number wins each time, three survive, and the one with the smallest digit outside falls to the same argument as before. That leaves two. One has b and c in the weighted slots and a beside c. The other has c and b in the weighted slots and a beside b.

6:52b times c, and c times b. The big pieces agree, exactly as they did with numbers. So the small pieces decide: a times c against a times b. And c is the larger, so the first one wins. The largest digit goes outside as the multiplier. The other two go inside, larger one in the tens place. That is not a rule anyone announced. It is what was left when everything else had been eliminated.

7:25Try it cold on digits the argument has not seen. One, three and seven. Largest outside, so seven multiplies; the other two inside with the larger first, so thirty-one. Thirty-one times seven. That is two hundred and seventeen, and it is the biggest of the six. The runner-up, seventy-one times three, is two hundred and thirteen. Now three, five and nine. The rule says nine outside and fifty-three inside. Fifty-three times nine, which is four hundred and seventy-seven.

8:04This one is worth pausing on, because the two arrangements your hand wants to write are ninety-five times three and ninety-three times five. They come to two hundred and eighty-five, and four hundred and sixty-five. Both lose. Making the two-digit number as large as you possibly can is exactly the wrong instinct. The multiplier is a place too, and it multiplies everything. Every argument leans on something. This one leans on two things worth naming.

8:41The first is that a gain in the small piece can never pay for a loss in the big one. That sounds obvious and it is not free, so it gets checked: for digits under ten it holds every time, and the differences are never even close. The second is that all three digits are different. Watch what happens with a zero. Take zero, three and five. The rule says the zero goes in the ones place, where its weight is one, and that is the right place for it.

9:15But then the small piece is zero times something, which is nothing at all. And the small piece was what the contest was being settled by. So thirty times five and fifty times three both come to one hundred and fifty. The winner is no longer alone. Nothing has gone wrong; the argument simply runs out of things to compare. So here is what actually happened. Nothing was multiplied out until the answer had already been found.

9:49Group the arrangements so that one thing is held fixed, and within a group the comparison becomes obvious. Split what is left into a large part and a small part. Show the large parts are equal, and let the small part decide. Holding something fixed so that only one thing varies is not a trick for this puzzle. It is most of how comparison works. And the answer, in the end, is that the biggest digit belongs where it multiplies the most, which is outside.

10:24Six arrangements, and not one of them was multiplied out until the answer was already known.