i'm thinking on this...i'm new here so i don't quite know how to come back to your problem, but maybe i can trace it through this response. sorry i don't know what to tell you right now!
Hi all, I came across this problem and its solution:
There are 5 people A-E. They have alot of oranges. First, A takes 1 orange from the basket, and 1/5 of the remaining. Next, B takes 1 orange, and 1/5 of the remainder again. And So on. What is the least number of oranges that they have?
Basically the solution is 5^5-4.
I came up with a general case whereby
There are n people. They have alot of oranges. First, A takes 1 orange from the basket, and 1/n of the remaining. Next, B takes 1 orange, and 1/n of the remainder again. And So on. What is the least number of oranges that they have?
And the solution should be n^n - (n-1)
I understand the n^n but as for the n-1, I couldn't figure out the logic in it. Could anyone help me with this?
(My apologies if this thread is in the wrong topic)
What? No monkey to give an orange to?
Go here: The On-Line Encyclopedia of Integer Sequences™ (OEIS™)
enter sequence A006091
IF you can call this a "looper formula":
x = n^n - n + 1
DOWHILE (x-1) / n = INTEGER
x = x-1 : x = x - x/n
For similar problems, google "monkey and coconuts"