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Number Of Coconuts In The Pile

**What was the problem?**
Absolutely no need to overthink on the **extra details** given there. Just for a moment, we assume the number of coconuts in the community pile is **divisible** by **10,9,8,7,6,5,4,3,2,1.**
Such a number in mathematics is called as **LCM.** And **LCM** in this case is** 2520.** Since each time 1 coconut was falling short of equal distribution there must be **2519** coconut in the pile initially. Let's verify the fact for all **10 distributions** tried by **10 people.**Each time monkey kills 1 person & number of persons among which coconuts to be distributed decreases by 1 each time.
1st Man - 2519/10 = 251 x 10 + 9.....1 short to equal distribution among 10.
2nd Man - 2519/9 = 279 X 9 + 8.......1 short to equal distribution among 9.
3rd Man - 2519/8 = 314 x 8 + 7........1 short to equal distribution among 8.

4th Man - 2519/7 = 359 x 7 + 6........1 short to equal distribution among 7.
5th Man - 2519/6 = 419 x 6 + 5........1 short to equal distribution among 6.

6th Man - 2519/5 = 503 x 5 + 4........1 short to equal distribution among 5.
7th Man - 2519/4 = 629 x 4 + 3........1 short to equal distribution among 4.

8th Man - 2519/3 = 839 x 3 + 2........1 short to equal distribution among 3.
9th Man - 2519/2 = 1259 x 2 + 1.......1 short to equal distribution among 2.

10th Man - 2519/1 = 2519..............Can take all of them

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