A woman took a certain number of eggs to the market and sold some of them.

Next day the number left over was doubled and she sold the same number as the previous day.

On the third day the remainder was tripled, on the fourth day quadrupled, and on the fifth day quintupled.

Each day she sold exactly the same number and finally disposed of her entire stock.

What is the smallest number of eggs she could have taken on the first day and how many did she sell daily?

Let the number of eggs sold each day be $x$ and the number of eggs remaining after the first day be $a_1$. Day 1: Initial eggs $= x + a_1$ Day 2: Leftover doubled, then sold $x$ $a_2 = 2a_1 - x$ Day 3: Leftover tripled, then sold $x$ $a_3 = 3a_2 - x$ Day 4: Leftover quadrupled, then sold $x$ $a_4 = 4a_3 - x$ Day 5: Leftover quintupled and all sold $5a_4 - x = 0$ $\Rightarrow x = 5a_4$ Working backward From $x = 5a_4$ $a_4 = \dfrac{x}{5}$ From $a_4 = 4a_3 - x$ $\dfrac{x}{5} = 4a_3 - x$ $4a_3 = \dfrac{6x}{5}$ $a_3 = \dfrac{3x}{10}$ From $a_3 = 3a_2 - x$ $\dfrac{3x}{10} = 3a_2 - x$ $3a_2 = \dfrac{13x}{10}$ $a_2 = \dfrac{13x}{30}$ From $a_2 = 2a_1 - x$ $\dfrac{13x}{30} = 2a_1 - x$ $2a_1 = \dfrac{43x}{30}$ $a_1 = \dfrac{43x}{60}$ Initial number of eggs Initial eggs $= a_1 + x$ $= \dfrac{43x}{60} + x$ $= \dfrac{103x}{60}$ For this to be an integer, smallest possible value is $x = 60$ So initial eggs $= \dfrac{103 \times 60}{60} = 103$Initial number of eggs $= 103$ Number sold daily $= 60$