• Jun 24th 2010, 06:40 AM
spaarky
Abel, Bob, Conan, Dave and Elijah divided a certain number of marbles amongst themselves in the following way: Abel took 1 marble and 20% of the remaining marbles, then Bob took 1 marble and 20% of the remaining marbles. Conan, Dave and Elijah did the same. At least how many marbles were there in tbe beginning?

I have two answers but do not know which is correct. Is it a matter of how you interpret the 20% of the remaining ? Math is meant to be black and white but when different ppl say different things ... confusing

3121 and 10

[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + R))))) units not drawn to scale
[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + 5/4 (1 + R)))) units not drawn to scale
[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + 5/4 (1 + R)) units not drawn to scale

[1][ ][ ][ ][ ][ ] --> 1 + 5/4 (1 + R)
[1][][][][][] --> 1 + R

As R is a multiple of 5, try till you get the first expression as a whole number. Solve using Microsoft Office - Excel, the first whole number is 3121.

There were 3121 marbles in the beginning.

each person has 1 marble plus 20% of the 'remainder'
--> There are 5 marbles + 5x20% of the 'remainder'
--> 100% of the remainder is used / split across the 5 people

So, what is the smallest number that satisfies these conditions
- 100% of the Remainder / 5 is a positive whole number (cannot have a fraction of a marble nor a negative marble)
- the Remainder is greater than zero. If not than, 0/5 = 0 producing a whole number. I don't think they are looking for this answer however. I would think that the Remainder needs to be greater than 0. With this assumption, the Remainder needs to be a multiple of 5.

The smallest multiple of 5 is 5.
--> 5 marbles + 5 marbles = 10 marbles

The smallest number of marbles is 10.
• Jun 24th 2010, 09:36 AM
SpringFan25
i think it wants answer 1 as well.

edit typo fixed
• Jun 24th 2010, 11:04 AM
Your first answer is correct: the lowest number you can start with is 3121. This leaves 1020 marbles at the end, which is $4^5-4$. In fact, all solutions will leave a number of marbles in the form $4^5n-4,\;n=1,\; 2,\; 3,\; ...$ I'll leave it to someone else (or you!) to prove it!