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So this question came up in my exam and luckily I had time so I used trial and error to solve it however I am looking for a rigorous equation to solve it.

The question statement is that there is an odd number of stones along a road and each of them placed at $10\mathrm{m}$ from each other. A man starts at one end and picks up one stone and puts on the center stone. He then goes back to the next stone from the end and picks it up and puts it on the center stone. He repeats the process until all the stones are on the center stone. If the total distance covered is $3\mathrm{km}$ then find the number of stones. I figured by trial and error that the answer was $25$ stones.

The only thing that I was able to come up with is if we assume that the distance between each stone is $d$ and thus $d=10\mathrm{m}$. Also, let $n$ be the number of stones. Now we get for $25$ stones,

Total distance covered = $$\Biggl(4\cdot\frac{\bigl(n-3\bigl)}{2}\cdot d + 4\cdot\frac{\bigl(n-5\bigl)}{2}\cdot d + 4\cdot\frac{\bigl(n-7\bigl)}{2}\cdot d +\ldots 4\cdot\frac{\bigl(n-23\bigl)}{2}\cdot d + 3\cdot\frac{\bigl(n-1\bigl)}{2}\cdot d \Biggl)$$ For $n=25$ I get the distance as $3000\mathrm{m}$ . I need some help to define the equation for the number of stones, however, given the distance. Any help would be appreciated.

Prakhar Nagpal
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1 Answers1

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Assuming there are $2n+1$ stones, there will be $n$ stones each on right side and left side of the middle one. It is also given that there are $n$ intervals each $10$m on both the sides.

If the man collects the stones from first to $n$ stones and drops it around the middle stone:

The distance covered in collecting all the stones on the left side: $$d_{\text{left}} = 2[10(1) + 10(2) + 10(3) +… + 10(n-1)] + 10(n)\tag 1$$ Why? Because the man was initially at the extreme left, then to take the $n^{\text{th}}$ stone, he will travel one way whereas for the other stones he travels two ways.

Now the person is at the middle stone. The process is almost the same except for the fact that here he will now cover distance both ways in collecting all the stones.

Hence the distance covered on the right side: $$d_{\text{right}} = 2[10(1) + 10(2) + 10(3) +… + 10(n-1) + 10(n)]\tag 2$$

Hope you can take it from here. See also this.