# Part-Time Carpenter

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## Classic Problem

A carpenter agrees to work, on the condition that:

- he is paid $\pounds 2$ for every day he works
- he forfeits $\pounds 3$ for every day he does not work.

At the end of $30$ days he finds he has paid out exactly as much as he has received.

## Solution

$18$ days.

## Proof

Let $x$ be the number of days he worked.

Then:

\(\ds 2 x\) | \(=\) | \(\ds 3 \paren {30 - x}\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds 5 x\) | \(=\) | \(\ds 90\) | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds 18\) |

Checking our work:

For $18$ days and at $\pounds 2$ per day, he earns $\pounds 2 \times 18 = \pounds 36$.

For $12$ days, forfeiting $\pounds 3$ per day, he forfeits $\pounds 3 \times 12 = \pounds 36$.

$\blacksquare$

## Sources

- 1484: Nicolas Chuquet:
*Triparty en la Science des Nombres* - 1976: Howard Eves:
*Introduction to the History of Mathematics*(4th ed.): p. $235$ - 1992: David Wells:
*Curious and Interesting Puzzles*... (previous) ... (next): The Nuns in their Cells: $96$