And for those who find the riddle above too complicated, I reiterate the magic trick from the circuit breakers:
Here is a magic trick:
You ask a person to think of a number between 1 and 100.
You give them a set of cards with numbers and ask the person to tell you on which cards their number appears.
Now you can say what the number was.
How are those cards made? (which numbers are written on each card?)
How many cards do you need?
How do you know what their number was?
Hidden:
7 Cards.
card A = #1 thru 50
B = 1-25 and 51-75
C = 1-13, 26-38, 51-63 and 76-88
D = 1-7, 14-19, 26-32, 39-44, 51-57, 64-69, 76-82 and 89-94
E = 1,2,3 8,9,10 14,15,16 20,21,22 26,27,28 32,33,34 38,39,40 44,45,46 50,51,52 56,57,58 62,63,64 68,69,70 74,75,76 80,81,82 86,87,88 92,93,94 98,99,100
F = from every group of 3 number listed above, put two on this card. Additionally, put two of every set of 3 numbers "between" those sets,
G = Every other number from card F.
Any given set of cards will only have one number in common.
7 Cards.
card A = #1 thru 50
B = 1-25 and 51-75
C = 1-13, 26-38, 51-63 and 76-88
D = 1-7, 14-19, 26-32, 39-44, 51-57, 64-69, 76-82 and 89-94
E = 1,2,3 8,9,10 14,15,16 20,21,22 26,27,28 32,33,34 38,39,40 44,45,46 50,51,52 56,57,58 62,63,64 68,69,70 74,75,76 80,81,82 86,87,88 92,93,94 98,99,100
F = from every group of 3 number listed above, put two on this card. Additionally, put two of every set of 3 numbers "between" those sets,
G = Every other number from card F.
Any given set of cards will only have one number in common.
The number of cards is right...
but how will you find out what the number was?
I have a much easier solution for designing the cards that will also tell you imediately what the number was...
This one’s serious....
The warden meets with 23 new prisoners when they arrive to prison. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison is a switch room, which contains two light switches labeled A and B. Both are in the “off” position now. The switches are not connected to anything.
All day long, I will be leading prisoners randomly, one at a time, into the switch room. Every time you’re there, you must flip one switch and you’ll then be led back to your cell.
There’s no limit how many times a prisoner will visit the room, and no particular order which I will follow. One might visit the room 3 times before another would be there even once, but with time, all prisoners will have visited the room the same amount of times.
At any given time, any prisoner may yell “all have visited the room!”
If that will be true, you’ll all be freed. If even one prisoner hasn’t been to the room yet, you’ll all be shot.
I will not purposefully hold back one prisoner from entering the room.”
What plan can they come up with to ensure that they’re freed?
Hidden:
Just came here, and would not have guessed without a hint but:
Any person who comes in for the first time switches on A only the leader closes A when he comes in. If A is on when u come in for the first time you wait for a time a is off and then you switch it on. Leader keeps count of all the times he switches it off.
Everyone’s else constantly flips switch b.
Maybe extra credit can help if she remembers how the trick was done...
It was a quite well-known trick...
I hardly remember the day of the week, hard to remember things from a couple of decades ago. But nice to remember that we once kept ourselves entertained with real live humans instead of screens
I believe the first digit of each card it’s written on will add up to the answer but now I gotta see how many cards and what those first digits should be
I believe the first digit of each card it’s written on will add up to the answer but now I gotta see how many cards and what those first digits should be
Yes!
And tamari: what you wrote was right... it had to do with the first # on each card...
7 Cards.
card A = #1 thru 50
B = 1-25 and 51-75
C = 1-13, 26-38, 51-63 and 76-88
D = 1-7, 14-19, 26-32, 39-44, 51-57, 64-69, 76-82 and 89-94
E = 1,2,3 8,9,10 14,15,16 20,21,22 26,27,28 32,33,34 38,39,40 44,45,46 50,51,52 56,57,58 62,63,64 68,69,70 74,75,76 80,81,82 86,87,88 92,93,94 98,99,100
F = from every group of 3 number listed above, put two on this card. Additionally, put two of every set of 3 numbers "between" those sets,
G = Every other number from card F.
Any given set of cards will only have one number in common.
