Counterfeit Coin problem with a twist?
We all know the counterfeit coin problem with 7 real coins & one fake one.
n coins = x weighings
2 coins = 1 weighing
3 coins = 1 weighing
4 coins = 2 weighings
5 coins = 2 weighings
6 coins = 2 weighings
7 coins = 2 weighings
8 coins = 2 weighings
9 coins = 2 weighings
10 coins = 3 weighings
11 coins = 3 weighings
12 coins = 3 weighings
etc.
Well I get the problem and I sort of have the answer to it.
n^2 = k & the amount of digits k has = x weighs
How do I explain that in one equation/math sentence?
like n^2 x 3^3 or something like that…
Help please. Thanxx =]
I know how to solve it.
I just need to know how to put that the number of digits n^2 has is the amount of weighings or x
n/2^k =< 2
think about how you actually perform the weighings.
You divide the coins in half, and in half again. that first weighing tells you which half has the odd coin. You repeat until you have two coins left.