eli5

gotcha. thanks

I like that puzzle

at first it seems kinda like one of those arbitrary letters have a certain value and then do the same function each time but it's cleverer than that.

Well worked out.

My only problem with a puzzle like that is that there isnt really a "right answer". But I appreciate the idea of "can you work out a non-straightforward answer".

In general I agree with you. What's next in the sequence 2, 4, 8? etc questions annoy me.

For this one even though there's no single right answer, I'm certain that you couldn't come up with a better answer (defined by polling a bunch of people)

This one isn't a logic puzzle, per se, but it can be a pleasant way to pass a bit of time.

Find integers (all greater than 1) a, b, c, d, e, f, g such that the following are true:

a^2 × b × c^2 × g = 5,100

a × b^2 × e × f^2 = 33,462

a × c^2 × d^3 = 17,150

a^3 × b^3 × c × d × e^2 = 914,760

Where ^ denotes exponts, as usual.

Don't you need as many equations as variables to get definite answers? It's been a long time since I took an algebra class, but I thought that was the general rule.

But this problem requires integers. Requiring integers greatly restricts things

Well I went through a bunch of overthinking it math before I started taking a simpler approach, and it seems pretty straight forward.

My approach was the same as yours, irtm, but I started at a different point. The given solution (which is, in effect, equivalent):

Enjoyed this thoroughly.

The previous puzzle and this next one were taken from the NSA's puzzle periodical, which doesn't seem to be active this year.

New puzzle
---------------------

At a work picnic, Todd announces a challenge to his coworkers. Bruce and Ava are selected to play first. Todd places $100 on a table and explains the game. Bruce and Ava will each draw a random card from a standard 52-card deck. Each will hold that card to his/her forehead for the other person to see, but neither can see his/her own card. The players may not communicate in any way. Bruce and Ava will each write down a guess for the color of his/her own card, i.e. red or black. If either one of them guesses correctly, they both win $50. If they are both incorrect, they lose. He gives Bruce and Ava five minutes to devise a strategy beforehand by which they can guarantee that they each walk away with the $50.

Bruce and Ava complete their game and Todd announces the second level of the game. He places $200 on the table. He tells four of his coworkers -- Emily, Charles, Doug and Fran—that they will play the same game, except this time guessing the suit of their own card, i.e. clubs, hearts, diamonds or spades. Again, Todd has the four players draw cards and place them on their foreheads so that each player can see the other three players' cards, but not his/her own. Each player writes down a guess for the suit of his/her own card. If at least one of them guesses correctly, they each win $50. There is no communication while the game is in progress, but they have five minutes to devise a strategy beforehand by which they can be guaranteed to walk away with $50 each.

For each level of play – 2 players or 4 players– how can the players ensure that someone in the group always guesses correctly?


If, by some chance, this one has been in this thread already, don't look back.

2 players:

4 players:

2 players:

Lattimer's solution >>>>> mine

I've seen a similar puzzle with two people. Never seen a version of four. I imagine it's kind of the same concept.

Gabe, it could be argued that your solution involves a type of communication during the process. My initial plan was essentially the same.

I know that puzzle but without cards and with a general solution required for n people.

Howard Lederer said it on the PokerCast

He also told a puzzle which went something like this:

"Why are most manhole covers round?"

What I always heard about manhole covers was

Finally looked at the 4 person solution. Never would have got there.