Hard Riddle #7

Because the solution is naturally expressed in terms of logarithm... I don't know what else to tell you.

@polylogarithmique said in #11:

Because the solution is naturally expressed in terms of logarithm... I don't know what else to tell you.
...
Can you tell me why this solution works?

Label the wrappers 1-50, for each wrapper number n take out n coins and add to one big pile of coins, weigh the pile and see how much lighter the final reading is than expected

Hence 1 weighing

@CalbernandHowbe said in #13:

Label the wrappers 1-50, for each wrapper number n take out n coins and add to one big pile of coins, weigh the pile and see how much lighter the final reading is than expected
Hum do we know that every coin from the wrong wrapper are fake coins? Or only some of them?

it says in the q

It says "There is one wrapper which has fake pennies,". If it has 48 true pennies and two fake pennies, it still "has fake pennies".

https://www.youtube.com/watch?v=mNgshmsXpCc

By the way, what is the sum of all numbers (1-36) on a roulette wheel ? Use the good old Gauss method.

P.S. I only mentioned this because I tried to solve the original riddle three times with different "halving the stack methods". My wrong solution was always 6. Hence 6-6-6. Pretty devilish.

@Bishop1964 said in #19:

By the way, what is the sum of all numbers (1-36) on a roulette wheel ? Use the good old Gauss method.
(1 + 36) x 18 = 540 + 126 = 666
AYO WHAT TH-