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.
(In reply to
re: So close yet so far? by Charlie)
For all solutions given {a,b}
{a = (7n+1)(7^7)6, b = 6(6^6n+6665)}
When n = 0, {a = 823537, b = 39990}
Since:
((7(n+1)+1)(7^7)6)((7n+1)(7^7)6) = 7(7^7) = 7^8
6(6^6(n+1)+6665)6(6^6n+6665) = 6(6^6) = 6^7
the common differences are 5764801 and 279936 respectively.
b is also 6^7n+1/7(6^76). This is the generalised share of all stepchildren.
As to consanguine children, number them 1 to 7; then the generalised share of the kth child is 6^(k1)*7^(8k)*n+6^(k1)*7^(7k)1, for all estates compliant with the problem, having a minimum when n=0.
Algebraic proof.
Let the original amount be k.
The consanguineous shares are then:
1st (k1)/7
2nd 1/7*(6((k1)/7)1) = 1/49*(6k13)
3rd 1/7*(6/49(6k13)1) = 1/343*(36k127)
4th 1/7*(6/343(36k127)1) = 1/2401*(216 k1105)
5th 1/7*(6/2401*(216 k1105)1) = 1/16807*(1296k9031)
6th 1/7*(6/16807*(1296k9031)1) = 1/117649*(7776 k70993)
7th 1/7*(6/117649*(7776 k70993)1) = 1/823543*(46656k543607)
Note that 1,13,127,1105...etc. =7^n6^n; the other integers are powers of 6 and 7, a key feature of the solution.
Now for stepchildren, let each receive x:
7x=6/7*(6/117649*(7776k70993)1)
x = (279936k3261642)/5764801
(7^8*x)+6*7^7= 6^7k+6^8
7^7(7x+6) = 6^7(k+6)
k = 7^8n+7^76, x = 6(6^6n+1/7*(6^61))
Let n=0
k = 823537
x = 6665*6 = 39990
Nice problem.
Edited on May 28, 2014, 4:23 pm

Posted by broll
on 20140528 09:12:04 