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