Henry Ernest Dudeney/Puzzles and Curious Problems/149 - Sheep Stealing/Solution

From ProofWiki
Jump to navigation Jump to search

Puzzles and Curious Problems by Henry Ernest Dudeney: $149$

Sheep Stealing
Some sheep stealers made a raid and carried off one-third of the flock of sheep, and one-third of a sheep.
Another party stole one-fourth of what remained, and one-fourth of a sheep.
Then a third party of raiders carried off one-fifth of the remainder and three-fifths of a sheep,
leaving $409$ behind.
What was the number of sheep in the flock?


Solution

The flock originally contained $1025$ sheep.


Proof

Let $n$ be the number of sheep in the flock.

Let $n_1$ and $n_2$ be the numbers remaining after the first and second raids respectively.

We have:

\(\ds 409\) \(=\) \(\ds n_2 - \paren {\dfrac {n_2} 5 + \dfrac 3 5}\) Then a third party of raiders carried off one-fifth of the remainder and three-fifths of a sheep, leaving $409$ behind.
\(\ds \) \(=\) \(\ds \dfrac {4 n_2 - 3} 5\) simplification
\(\ds \leadsto \ \ \) \(\ds n_2\) \(=\) \(\ds 512\) simplification


\(\ds 512\) \(=\) \(\ds n_1 - \paren {\dfrac {n_1} 4 + \dfrac 1 4}\) Another party stole one-fourth of what remained, and one-fourth of a sheep.
\(\ds \) \(=\) \(\ds \dfrac {3 n_2 - 1} 4\) simplification
\(\ds \leadsto \ \ \) \(\ds n_1\) \(=\) \(\ds 683\) simplification


\(\ds 683\) \(=\) \(\ds n - \paren {\dfrac n 3 + \dfrac 1 3}\) Some sheep stealers made a raid and carried off one-third of the flock of sheep, and one-third of a sheep.
\(\ds \) \(=\) \(\ds \dfrac {2 n_1 - 1} 3\) simplification
\(\ds \leadsto \ \ \) \(\ds n\) \(=\) \(\ds 1025\) simplification

$\blacksquare$


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Henry_Ernest_Dudeney/Puzzles_and_Curious_Problems/149_-_Sheep_Stealing/Solution&oldid=561759"