Hard Riddle #7

This one requires some out of the box thinking.

Suppose I have 50 coin wrappers, each with 50 coins. Each penny weighs 2.5 g. There is one wrapper which has fake pennies, each of the fake pennies weigh 0.5g less than an actual penny. The only tool you have is a weighing scale. What is the lowest amount of times that you would have to weigh the pennies to identify which of the wrappers has the counterfeit coins?

Lowest possible would be two.
Set aside two rolls, weigh 24 rolls against 24 rolls. If they balance weigh remaining two rolls against each other,lighter roll is counterfeit.
Alternatively set aside two rolls,weigh 24 against 24,discard heavier coins and two set aside,weigh 12 against 12 discard heavier,weigh 6 against 6 discard heavier, weight 3 against 3 discard heavier.Of remaining 3 weigh 1 against 1,if they balance third roll is counterfeit if they do not lighter roll is counterfeit
So two minimum,five maximum

@Dukedog Sorry, I should have clarified. The weighing scale isn't a balancing one, it is a digital weighing scale. Also, the question is what is the lowest amount of times in which you will always know which wrapper has the counterfeit coins. The answer should not have a maximum and a minimum, if it does, than the final answer is said to be the maximum.

We can ignore the wrappers and just say there are 50 coins, 1 which is fake.
Let us form 2 groups of 24 pennies and one group of 2 pennies.

Possibilities:
If the two groups of pennies are equal, we weigh the 2 pennies against each other to see which one is the fake.
If one group of pennies is larger than the other, we have 24 pennies leftover.

This time, we can split the pennies into 3 groups of 8 pennies.
We weigh two groups of pennies.

Possibilities:
If the two groups of pennies are equal, we turn to the other group of 8 pennies.
If not, we turn to the group that was lighter.

We split the 8 pennies into 2 groups of 4 pennies and weigh them.
One group is lighter than the other.

We split the 4 pennies into 2 groups of 2 pennies.
One group is lighter than the other.

We split the 2 pennies into 2 groups of 1 penny and weigh them.
Counterfeit coin is found.

Five steps.

@InkyDarkBird Incorrect. You are taking the wrong approach, though you are right in saying that you can ignore the wrapper as all the wrappers have the same mass. I will say it again, this one requires some out of the box thinking. In my opinion, this is the hardest riddle that I have asked so far because it requires the understanding of a math trick that is rarely used.

@TakeThePawnOrLose STOP GIVING US MATH QUESTIONS THESE ARE NOT RIDDLES JUST FIND SOME ON THE INTERNET

@your_good_side GUESS WHAT, YOU DON'T NEED TO EVEN READ THESE. YOU COULD JUST IGNORE THEM BECAUSE LET'S BE HONEST, YOU DON'T HAVE THE MENTAL ABILITY TO SOLVE THEM.

I give up.

I think something like 2[log(50)/log(2)] should do it.

Edit: actually just [log(50)/log(2)] because we already know the weight of the coins.

@polylogarithmique said in #9:

Edit: actually just [log(50)/log(2)] because we already know the weight of the coins.
Why are you using logarithms?