Part-Time Carpenter
Jump to navigation
Jump to search
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: $96$