A certain rich man had no sons, but several daughters. When he died, most of his estate was divided in the usual ways, except for his collection of pearls.

The will stated that that the first daughter was to receive 1 pearl, plus 1/7th of the remaining pearls. The second daughter was to receive 2 pearls, plus 1/7th of the remaining pearls. The third daughter was to receive 3 pearls, plus 1/7th of the remaining pearls, and so on.

The daughters, of course, went to a judge and protested this confusing and unfair method of dividing up the pearls. The judge looked over the will, and looked over the pearls, and was able to point out that each daughter would receive exactly the same amount of pearls, thus making the method of division of fair.

How many pearls did the rich man leave his daughters?
For that matter, how many daughters did the rich man have?

I'll assume that the problem has a solution as the judge says. I'll then assume that daughter number d (d>=1) gets p pearls of the original P pearls, so

p = d + (P-(d-1)p-d)/7

which gives

P = (p-6)d +6p

As p and P obviously should be constant and definitely not dependant on d we see that we have to chose p = 6 and the same equation says that then P = 36.

He left 36 pearls (6 to each daughter) which would make it 6 daughters.

Although, _if_ there is such a thing as negative pearls he could have infinitely many daughters. Daughter number 7 would look at the pile of zero pearls and remove 7 of them leaving 7 negative pearls. She would then receive 1/7 of the 7 negative pearls, which is one negative pearl, leaving her with 6 pearls, just as her 6 previous sisters got. You see the first formula holds for all d>=1 so daughter 8, 9 and so on would get 6 pearls each also.

/Tomas