A woman took a certain number of eggs to the market and sold
some of them.
The next day, through the industry of her hens, 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
sales 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 stock.
What is the smallest number of eggs she could have taken to
market the first day, and how many did she sell daily? Note
that the answer is not zero.
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Answer Posted / guest
She took 103 eggs to market on the first day and sold 60
eggs everyday.
Let's assume that she had N eggs on the first day and she
sold X eggs everyday. Putting down the given information in
the table as follow.
Days Eggs at the start of the day Eggs Sold Eggs Remaining
Day 1 N X N-X
Day 2 2N-2X X 2N-3X
Day 3 6N-9X X 6N-10X
Day 4 24N-40X X 24N-41X
Day 5 120N-205X X 120N-206X
It is given that she disposed of her entire stock on the
fifth day. But from the table above, the number of eggs
remaining are (120N-206X). Hence,
120N - 206X = 0
120N = 206X
60N = 103X
The smallest value of N and X must be 103 and 60
respectively. Hence, she took 103 eggs to market on the
first day and sold 60 eggs everyday.
| Is This Answer Correct ? | 40 Yes | 4 No |
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