Mr. Saunders has recently passed away.
In his will, he has divided his money among his 7 children, 7 stepchildren and a manager.
Mr. Sanders’ will stipulated:
 First give $1 to manager, and 1/7th of the remaining to the eldest child.
 Now give $1 to the manager and 1/7th of the remaining to second oldest child and so on.....
 After giving the money to the seventh (youngest) child, divide the remaining money among the seven stepchildren equally.
Knowing that:
 The ages of each of the seven children are different, and:
 Each of the 15 individuals received a positive integer number of dollars
What is the minimum amount of money that Mr. Saunders had left behind?
Extra Challenge: A non computer program aided solution.
I came up with the same solution. Note that starting amount, 823537 = 7^7  6.
I've been trying to write up an analytical explanation but I keep getting stuck. Maybe someone here can shed some insight.
It's tempting to think it should be something like this: At the outset, you need an amount of money that is equal to 6 mod 7, since you need to take away a dollar and have what remains be divisible by 7. But then after you remove 1/7 of this, you need what remains to once again be 6 mod 7, so the amount you start with must actually be 6 mod 7^2 in order to pay the manager twice and two children.
Similarly, in order to pay the manager three times and three of the children, you'd need to start with an amount that is 6 mod 7^3...
So since you have to pay the manager 7 times and 7 children, you need to start with an amount that is 6 mod 7^7. The minimum such positive integer amount is 7^7  6, which turns out to be the correct solution.
But this reasoning doesn't work, because after the 7th child we don't need an amount that is 6 mod 7 because we're not paying the manager again before dividing what remains among the 7 stepchildren. How is it then that faulty reasoning leads to the correct answer anyway?
Surely it's not entirely a coincidence, but I'm getting mixed up trying to make sense of it. Furthermore I observed that if the number of children N is 3 or 5 or 7 or 9, then 7^N  6 works as a solution; but if N is 2 or 4 or 6 or 8, then 7^N 6 doesn't work, so it's not a general solution.

Posted by tomarken
on 20140527 17:03:58 