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 have a favorite puzzle collecting dust in your attic, find me on Twitter.

## Riddler Express

From Josh Kaplan, a classic coin puzzle:

You have nine gold coins, but one isn’t pure. One has been minted with a cheap alloy, and is known to be heavier than the others. You have a simple balance scale. How do you determine the impure coin with only *two weighings*?

## Riddler Classic

Also from Josh, the mystery of the ersatz coin deepens:

You have 12 gold coins — or so you think! One is fake and is known to have a different weight than the others. It could be heavier or lighter; you only know it’s wrong. Using the same simple balance scale, how do you determine the incorrect coin, and whether it is heavier or lighter, in only *three weighings*?

## Solution to last week’s Riddler Express

Congratulations to 👏 Rachel Cass 👏 of Medford, Massachusetts, winner of last week’s Express puzzle!

In a legislature, there are 100 politicians. Some of them are honest and some of them are crooked. If you choose any two of them at random, at least one will be crooked. How many honest politicians are there?

There is only **one**. If there were more than one, you’d run some risk of choosing two honest politicians randomly from the legislature, which would violate the terms of the problem. But we also know that *some* of the legislature is honest, so there can’t be zero honest politicians.

## Solution to last week’s Riddler Classic

Congratulations to 👏 Amy Leblang 👏 of Wayland, Massachusetts, winner of last week’s Classic puzzle!

You have four ropes and a lighter. Each rope takes exactly an hour to burn from end to end, but they all burn at a nonconstant rate. You can light the ropes at their ends at any time you’d like. If you’re strategic in your pyromania, how many different specific lengths of time can you accurately measure?

You can measure **23 different lengths**. Here’s an explanation, adapted from the puzzle’s submitter, Peter Calhoun:

The first trick is realizing that you can record 30 minutes by burning both ends of one rope. Since you know it takes an hour to burn through a rope from one end to the other, once the two burns meet you know each will have burned through 30 minutes of rope (even though they might not meet at the center of the rope thanks to the nonconstant burn rates). You also have to consider burning the ends of the ropes at different times. For example, to measure 45 minutes, you can burn both ends of the first rope and one end of the second rope. After 30 minutes have passed, you can burn the other end of the second rope, making 45 minutes.

The calculation gets very tricky as we add ropes. Below are the possible lengths of time (in minutes, not including zero):

One rope: 30 and 60

Two ropes: 30, 45, 60, 90 and 120

Three ropes: 30, 45, 52.5, 60, 67.5, 75, 90, 105, 120, 150 and 180

Four ropes: 30, 45, 52.5, 56.5, 60, 67.5, 71.25, 75, 78.75, 82.5, 86.25, 90, 97.5, 105, 112.5, 120, 127.5, 135, 150, 165, 180, 210 and 240

Some of these time points are difficult to determine. For example, below are the many steps it takes to measure precisely 71.25 minutes with four ropes. (Let \(r_i\) represent rope \(i\).)

- Light both ends of \(r_1\), one end of \(r_2\), and one end of \(r_3\)
- 30 minutes pass (\(r_1\) burned through)
- Light the other end of \(r_2\) and one end of \(r_4\)
- 15 minutes pass (\(r_2\) burned through)
- Light the other end of \(r_3\)
- 7.5 minutes pass (\(r_3\) burned through)
- Light the other end of \(r_4\)
- 18.75 minutes pass (\(r_4\) burned through)

The total time passed is 71.25 minutes (=30+15+7.5+18.75).

Solver Henry Maxfield plotted the times you can measure (in hours) using six, five, four, and so on down, different ropes:

And solver Thomas Conroy wrote a program in C++ to find the lengths of time, which he was kind enough to post to GitHub.

Finally, solver Russell Yu proposed the following general solution for the number of lengths of time, *T*, you can measure with *N* ropes:

\begin{equation}T = 3 \cdot 2 ^ {N – 1} – 1\end{equation}

## Want to submit a riddle?

Email me at oliver.roeder@fivethirtyeight.com.