Think of a Number/Examples/Bachet/3
Jump to navigation
Jump to search
Classic Problem
- The first person takes a number of counters greater than $5$.
- The second person takes $3$ times as many.
- The first person gives $5$ counters to the second.
- The second person then gives the first $3$ times as many as the first person holds in his hand.
- How many counters has the second person got in his hand?
Solution
$20$
Proof
Let $A$ and $B$ denote the first and second person respectively.
Let $x$ be the number taken by $A$.
Then $B$ takes $3 x$ from $A$.
Then $B$ gives $5$ counters to $A$.
At this point, $A$ has $x = 5$ and $B$ has $3 x + 5$.
Then $A$ gives $3 \paren {x - 5}$ to $B$.
Hence $B$ now has $3 x + 5 - 3 \paren {x - 5}$
Thus, doing the algebra, we see:
- $3 x + 5 - 3 x + 3 \times 5 = 20$
$\blacksquare$
Sources
- 1612: Claude-Gaspar Bachet: Problèmes Plaisans et Delectables qui se font par les Nombres
- 1992: David Wells: Curious and Interesting Puzzles ... (previous) ... (next): Bachet: $107$