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Can You Outsmart Our Elementary School Math Problems?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-sized and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. If you need a hint, or if you have a favorite puzzle collecting dust in your attic, find me on Twitter.

Riddler Express

From Trevor Ferril, some cafeteria multiplication:

Two intelligent, honest students are sitting together at lunch one day when their math teacher hands them each a card. “Your cards each have an integer on them,” the teacher tells them. “The product of the two numbers is either 12, 15 or 18. The first to correctly guess the number on the other’s card wins.”

The first student looks at her card and says, “I don’t know what your number is.”

The second student looks at her card and says, “I don’t know what your number is, either.”

The first student then says, “Now I know your number.”

What number is on the loser’s card?

Submit your answer

Riddler Classic

From Josh and Laura Pasek, another elementary school (but not elementary in difficulty) math problem:

Consider the following game. In front of you is a stack of 10 cards printed with the numbers 0 through 9, one per card. The stack is shuffled and, sight unseen, you draw a number from the top. You look at the number and place it somewhere in the multiplication equation below. You then draw another number, look at it, and place it somewhere else in the equation. You do that two more times, until all four slots are filled. Once a digit is placed, it can’t be moved, and it can’t be drawn again because it’s no longer in the stack.

Your goal is to build a multiplication equation with the lowest possible product. What is the optimal strategy? And how much of this game is luck and how much is skill? In other words, how much does the expected product under the optimal strategy differ from simply placing the cards randomly?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Dan Mitchell 👏 of Minneapolis, winner of last week’s Express puzzle!

You and I agree to meet in our favorite park for a picnic. We each agree to arrive sometime between noon and 1 p.m., and we agree that whoever arrives first will wait for the other for up to 15 minutes before leaving to spend the day with a more punctual friend. If we each arrive at an independently random time during that hour, what are the chances we’ll actually have lunch together?

The chances are 7/16. This is a problem that can be solved geometrically, which makes it a lot more convenient to answer. Let’s start with a picture, with my possible arrival times on one axis and your possible arrival times on the other.

The shaded area is all the pairs of times that are within 15 minutes of one another — that is, when we successfully meet up to picnic! Now we just need to calculate its area.

Work from the outside in. The area of the square is its width times its height, or one hour-squared. The area of each unshaded triangle is one-half its base times its height, and each of those dimensions are three-quarters of an hour in our case. So each triangle has an area of (½)(¾)(¾). That leaves our shaded area as 1-2×((½)(¾)(¾)) = 7/16. In other words, there’s about a 44 percent chance we’ll have lunch together. I like those odds!

Solution to last week’s Riddler Classic

Congratulations to 👏 Luke Robinson 👏 of Oakwood, Ohio, winner of last week’s Classic puzzle!

The U.S. Chess Championship is in full swing, and last week I offered up two groups of chess puzzles to get us in the chess mood.

16 9
32 12
14 8
8 8
8 5

First, on a standard chessboard, what is the largest number of each piece (kings, knights, bishops, rooks and queens) that can be placed such that none of the pieces attack each other? Second, on a standard chessboard, what is the smallest number of each piece that can be placed such that every empty square is under attack? A summary of the solutions is in the table.

And here are examples of what those solutions could look like on the board, created with’s board editor. First, the largest number such that no piece attacks another:

Some of these solutions are fairly straightforward. For example, a knight, by rule, always attacks squares of the opposite color of the square on which it sits. Therefore, you can put knights on all the black squares and be sure none will attack another. Half the squares are black, so you can deploy 32 knights in this way. (You may need to buy a few extra chess sets to do this, however!) Also fairly straightforward are the rooks. Given their north-south, east-west attacks, you can just string them up the board’s main diagonal and they’ll all remain safe.

The solution for queens, on the other hand, is more intricate. You can pack in eight of them safely, but you’ve got to be precise about how you do it. In fact, for larger chess boards of size N-by-N, you can always pack in N queens. For a standard board, there are 92 different arrangements of queens that will work.

Second, here are examples of the smallest number of pieces such that each empty square is attacked:

Again, for some pieces, the arrangements are straightforward. The rook, for example, bites off chunks of the board horizontally and vertically, so as long as they don’t get in each other’s way, they can efficiently deploy their attack. Lining them up along the diagonal does that. The knights, however, with their L-shaped hopping, are trickier this time. It takes intricate trios of horsemen to successfully attack all the pesky squares in the corners.

And, once again, the queens might be the most interesting piece to think about. An extensive mathematical literature has sprung up around the queens problem. Whereas the rooks line up in a brutish diagonal phalanx to dominate the board, the five queens are snipers, and must be deployed wisely and intricately. Similar to the earlier problem, there are 91 arrangements that meet our criteria. Mathematicians have figured out the minimum number of queens needed to dominate boards of various sizes. You could also attack all 121 squares of an 11-by-11 board with just five queens!

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  1. Important small print: For you to be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday. Have a great weekend!

Oliver Roeder is a senior writer for FiveThirtyEight.