Richie Rich

Let $a_i$ denote the number of coins that the $i^\text{th}$ son receives. Let $x$ equal the number of coins that the rich man left behind. Note that since the first son receives $a_1$ coins, and that $a_1$ equals 1/5th of the remaining coins left after the first step, then there are 4/5 of the remaining coins left at the start of the second step, which is equal to $4a_1$ . Then, the second son receives $\frac{4a_1-1}{5}$ coins. This relation also applies to the other sons.

Then, clearly, $\frac{x-1}{5} = a_1 \\ \frac{4a_1-1}{5}=a_2 \\ \frac{4a_2-1}{5}=a_3 \\ \frac{4a_3-1}{5}=a_4 \\ \frac{4a_4-1}{5}=a_5$

Substitution into the equation $\frac{4a_4-1}{5}=a_5$ reveals that $\frac{256a_1-369}{625} = a_5$ , or that $256a_1-369 = 625a_5$ .

Coincidentally, $256+369=625$ so, when taken modulo 625, we get $a_1 \equiv 624 \pmod {625}$ makes $a_5$ an integer. Then, from $\frac{x-1}{5} = a_1$ we get that $x = \boxed{3121}$ . We must check, however, to see that 3121 allows each of $a_i$ to be an integer. It does work. 3121 is also the minimum number of coins that the rich man could leave behind, due to the way that $a_1$ was expressed mod 625.

Let the rich man have a total amount $n$ . The first son gets $\dfrac{n-1}{5}$ . The second son gets

$\Rightarrow \dfrac{\dfrac{4(n-1)}{5} - 1}{5} = \dfrac{4n - 9}{25}$ .

We can form a recurrence for the amount the $x^{\text{th}}$ son receives. The recurrence is as follows

$\Rightarrow f(1) = \dfrac{n-1}{5}$ and

$\Rightarrow f(x) = \dfrac{4(f(x-1)) - 1}{5}$

We can solve this recurrence and get a closed form. Then we find that the closed form is

$f(x) = n.4^{x-1}.5^{-x} + \left(\dfrac{4}{5}\right)^{x} - 1$

For the fifth, son, that is $x = 5$ , we get an amount $\dfrac{256n - 2101}{3125}$ after plugging $x = 5$ in the formula above.

Now, $\dfrac{256n - 2101}{3125}$ must be divisible by $5$ because this amount is divided equally to $5$ daughters, or in other words,

$\dfrac{256n - 2101}{15625}$ must be an integer. I found the smallest value of $n$ for which that expression is an integer with a small Python script.

It probably could have been done is a smarter way but right now I was too lazy to try to do that. The final answer was $\boxed{n = 3121}$ .

Let $y$ be the initial number of coins. Each daughter gets $\frac {4 (256y-2101)}{15625}$ coins. Of course if this is an integer, then all others will get an integer number of coins.

So it remains to solve $256 y = 2101 (mod 15625)$ . This gives $y= 3121 + 15625k$ for any integer $k$ . So $3121$ is the least possible.

Manager = 5 coins , then First son = 624 coins Second son = 499 coins Third son = 399 coins Forth son = 319 coins Fifth son = 255 coins Daughters = 204 each

This is a lengthy problem (for those who don't use computer scripts :P) Let's do it backwards...firstly...the number of coins the last son gets must be a multiple of 25 (as then only each daughter may get integral number of coins) Let the next son have x number of coins... thus (25n+1+x)/5=x (according to question) therefore on solving one may obtain that 25n+1 must be a multiple of 4. This happens only in case on a number ending with 75 (26,51,01 are not divisible by 4) Thus possible openings 75,175,275,375 and so on... also u may neglect multiples that do not begin with an even number(175,375 etc.)...as they add up to give a number which on adding 1 is not divisible by 4 Similarly...we may take into account only multiples of 4 as they dont give a number divisible by 4 in the third step. thus we shorten our list and reach upto 1275 which satisfies the given condition :D 1275+1=1276 1276/4=319 1276+319=1595 1595+1=1596 1596/4=399 1596+399=1995 1996/4=499 1996+499=2495 2496/4=624 2496+624=3120 hence the required answer 3121