Question 800856: 800 pennies, all in a row. All are 'heads' up.
Now, suppose that 800 people line up and.....
Person #1 turns over all of the pennies.
Person #2 turns over every 2nd penny, starting with penny #2.
Person #3 turns over every 3rd penny, starting with penny #3.
Person #4 turns over every 4th penny, starting with penny #4.
Etc...
After everyone has gone through the line, which pennies will show 'tails'?
Can't figure this out to save my life, please help. If you could include a short explanation of how to reach the answer, that would be great. I need to get my head around the process. :-) thanks
Answer by DrBeeee(684)   (Show Source):
You can put this solution on YOUR website! Let T(n) represent the position of a tail, T, as a function of n, I get
(1) T(n) = n^2 ; n = 1,2,3, ... 28
You get tails at position = (1,4,9,16,25,36,...,729,784).
Note that the space between the positions is
(2) space = (1,3,5,7,9,...,55).
You reason that the first person turns over the coin to a TAIL, and it never gets turned again so remains a TAIL.
The second person turns the second tail to heads and it remains heads. He then leaves the third coin as tails, turns the fourth etc..
The third person turns the third coin to heads, leaves the fourth as heads, fifth at tails, etc.
Now the fourth person turns the fourth coin from heads to TAILS, etc.
The important number to note is the "space" between the persons who leave a TAIL. So far we have a space of one between no-one and the first person, then a space of three from the first to fourth.
It turns out that the space is as given in (2), an sequence of odd numbers.
The solution logic results from the fact that space increases by two each time because the previous person skips turning the next number of coins equal to their position. For example, the eighth person will turn the eighth coin but skips the next 7. Thus the "chain" grows as people go through the line. For example, the 600th person in line only turns one coin!
|
|
|
