Henry Ernest Dudeney/Modern Puzzles/7 - Selling Eggs/Solution

by : $7$

 * Selling Eggs

Solution
She came to market on the first day with $103$ eggs, and sold $60$ every day.

Proof
Let $n$ be the original number of eggs taken to market.

Let $d$ be the number of eggs sold each day.

Let $r_1$, $r_2$, $r_3$ and $r_4$ be the remaining eggs after days $1$, $2$, $3$ and $4$.

On the first day, she sold $d = n - r_1$ eggs.

On the second day, she sold $d = 2 r_1 - r_2$ eggs.

On the third day, she sold $d = 3 r_2 - r_3$ eggs.

On the fourth day, she sold $d = 4 r_3 - r_4$ eggs.

On the fifth day, she sold $d = 5 r_4$ eggs.

So:

The smallest $n$ and $r_1$ that fit this equation are:

So:


 * She came to market on the first day with $103$ eggs.


 * She sold $60$ eggs per day.


 * At the end of the first day she had $43$ eggs remaining.


 * On the second day she started with $2 \times 43 = 86$ eggs, sold $60$ and had $26$ remaining.


 * On the third day she started with $3 \times 26 = 78$ eggs, sold $60$ and had $18$ remaining.


 * On the fourth day she started with $4 \times 18 = 72$ eggs, sold $60$ and had $12$ remaining.


 * On the fifth day she started with $5 \times 12 = 60$ eggs and sold them all.