How to know number of eggs in this problem?

Question

A women took a certain number of eggs to the market and sold some of them. The next day through her poultry industry, the number left over had been doubled, and she sold the same number as the previous day. On the third day the new remainder was tripled, and she sold the same number as before. On the fourth day the remainder was quadrupled, and her sale were the same as before. On the fifth day what had been left over were quintupled, yet she sold exactly the same as on all the previous occasions and so disposed of her entire stoke. What is the smallest number of eggs she could have taken to the market the first day and how many did she sell daily?
How to solve this question? Please describe your solution in steps.

Answer options are -

A. 110 and 50 B. 127 and 65 C. 100 and 60 D. 103 and 60

Answer 1

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

Let $e$ be the number she took and $s$ be the number she sold. On the way home after day 1 she has $e-s$.

The next day through her poultry industry, the number left over had been doubled, and she sold the same number as the previous day.

So on the way home after day 2 she has $2(e - s) - s = 2e - 3s$.

On the third day the new remainder was tripled, and she sold the same number as before.

So on the way home after day 3 she has $3(2e - 3s) - s = 6e - 10s$.

On the fourth day the remainder was quadrupled, and her sale were the same as before.

So on the way home after day 4 she has $4(6e - 10s) - s = 24e - 41s$.

On the fifth day what had been left over were quintupled,

So at the start of day 5 she has $5(24e - 41s) = 120e - 205s$.

yet she sold exactly the same as on all the previous occasions and so disposed of her entire stoke.

So $120e - 205s = s \Rightarrow 120e = 206s \Rightarrow e = \frac{103s}{60}.$

So $103s$ must be divisible by 60. 103 is prime, so $s$ must be divisible by 60. So the smallest positive $s$ is 60. She sold 60 eggs each day, and took 103 on the first day.

Answer 2

Why don't we start with

$x$ be the number of eggs left after selling $y$ eggs on the first day

Second day $:$ eggs doubled $\implies 2x$ remained $2x-y$ after selling $y$

Third day $:$ eggs tripled $3(2x-y)=6x-3y$ remaining $6x-3y-y=6x-4y$

Processing this way:

Ending at $\displaystyle 120x=86y\iff 60x=43y\implies\frac{43y}{60}=x$ which must be integer

But $(43,60)=1$

Answer 3

Low-tech no-algebra strategy: Since there are only four answer options, why not just tabulate what happens for each of them throughout the story? See which of them end up with 0 eggs after 5 days.