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 The Black Hole (Posted on 2004-02-18)
In a certain tribe, you have a certain amount of tribal offerings, at the start of year 1. However, at the start of each year (including this one), you must feed the black hole with a number of tribal offerings equal to the size of the black hole. On year 1, the black hole starts as size 1 and doubles each year that you pay the tribal offerings. (If it was 4, it's 8 now.) If you can't pay this cost, the island will explode in the middle of this year.

However, your workers are very industrious with investing, and always manage to double the number of tribal offerings that you had at the beginning of the year after paying the black hole.

For example, if you started with 4 offerings: (B = beginning of year before feeding the black hole, A = after you fed the black hole, E = end of year after your tribal offerings have doubled)

```--B-A-E
1|4 3 6
2|6 4 8
3|8 4 8
4|8 0 0
5|0
```
Since there wasn't enough to pay 16 tribal offerings, the island lasted 5 years.

How would you find the number of turns this island would last if you started with x tribal offerings?

 See The Solution Submitted by Gamer Rating: 3.2500 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 NO AMOUNT LASTS FOREVER | Comment 2 of 12 |
IF YOU START WITH X - YOU ARE LEFT WITH NOTHING AT THE BEGINNING OF YEAR X+1.

IT IS EVIDENT FROM A(N)=2*(A(N-1)-2^(N-2)) .Since
the gap between initial amount and the amount left after the offering grows exponentially no
initial x will be sufficient to last more than x
years.

a(n)=((((x-1)*2-2)*2-4)*2-8)*2 -2^(n-2)*2=0

iff n=x+1

 Posted by Ady TZIDON on 2004-02-18 11:08:39

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