Nov 10, 2009

Five selfish women, a monkey, and some coconuts (2 star)

Five women crash-land their airplane on a deserted island in the South Pacific. On their first day they gather as many coconuts as they can find into one big pile. They decide that, since it is getting dark, they will wait until the next day to divide the coconuts.

That night each woman took a turn watching for rescue searchers while the others slept. The first watcher got bored so she decided to divide the coconuts into five equal piles. When she did this, she found she had one remaining coconut. She gave this coconut to a monkey, took one of the piles, and hid it for herself. Then she jumbled up the four other piles into one big pile again.

To cut a long story short, each of the five selfish women ended up doing exactly the same thing. They each divided the coconuts into five equal piles and had one extra coconut left over, which they gave to the monkey. They each took one of the five piles and hid those coconuts. They each came back and jumbled up the remaining four piles into one big pile.

What is the smallest number of coconuts there could have been in the original pile?

P.S. Introducing stars...according to the difficulty of problem on the scale of 1 to 5.

3 comments:

1. int main()
{
float k;
int n;
for(n = 1; n < 10000; n = n + 1)
{
k = 0.8*(n - 1);
k = 0.8*(k - 1);
k = 0.8*(k - 1);
k = 0.8*(k - 1);
k = k - 1;

if(k - (int)k == 0.0)
if(((int)k)%4 == 0)
{
printf("Solution : %d",n);
break;
}
}
return 0;
}

Run this and you will get the answer.

2. kudos ...3121 score full marks :)
and kudos to KT too...