The Puzzle Thread

[leonid]

My take on cabinet puzzle

The first man's cabinet is labeled either AAA or AAO
The second man's cabinet is labeled either AAO or AOO

Case 1: The third man said yes -> his cabinet is labeled either OOO or AOO
Case 1.1: The first man's cabinet is labeled AAA

cabinet| 1 | 2 | 3 | 4 |
label  |AAA|   |   |   |
content|   |   |   |   |

cabinet| 1 | 2 | 3 | 4 |
label  |AAA|   |   |   |
content|AAO|   |   |   |
Cabinet #1 contains two apples and can't be AAA

cabinet| 1 | 2 | 3 | 4 |
label  |AAA|AAO|   |   |
content|AAO|AOO|   |   |
Cabinet #2 contains one apple and one orange and can't be AAO

cabinet| 1 | 2 | 3 | 4 |
label  |AAA|AAO|AOO|   |
content|AAO|AOO|OOO|   |
Cabinet #3 contains two oranges and can't be AOO

cabinet| 1 | 2 | 3 | 4 |
label  |AAA|AAO|AOO|OOO|
content|AAO|AOO|OOO|AAA|

Case 1.2: The first man's cabinet is labeled AAO

cabinet| 1 | 2 | 3 | 4 |
label  |AAO|   |   |   |
content|   |   |   |   |

cabinet| 1 | 2 | 3 | 4 |
label  |AAO|   |   |   |
content|AAA|   |   |   |
Cabinet #1 contains two apples and can't be AAO

cabinet| 1 | 2 | 3 | 4 |
label  |AAO|AOO|   |   |
content|AAA|AAO|   |   |
Cabinet #2's label can't be AAO
Cabinet #2 contains one apple and one orange

cabinet| 1 | 2 | 3 | 4 |
label  |AAO|AOO|OOO|   |
content|AAA|AAO|AOO|   |
Cabinet #3's label can't be AOO
Cabinet #3 contains two oranges

cabinet| 1 | 2 | 3 | 4 |
label  |AAO|AOO|OOO|AAA|
content|AAA|AAO|AOO|OOO|

We can't tell the cabinet's contents if the third person said "yes" because there are two possible outcomes


Case 2: The third man said no -> his cabinet is labeled either AAA or AAO

cabinet| 1 | 2 | 3 | 4 |
label  |   |AOO|   |   |
content|   |   |   |   |
Label AAA and AAO must be taken by cabinet #1 and #3 in some order,
So cabinet #2's label can only be AOO

cabinet| 1 | 2 | 3 | 4 |
label  |   |AOO|   |   |
content|   |AAO|   |   |
Cabinet #2 contains one apple and one orange and can't be AOO

cabinet| 1 | 2 | 3 | 4 |
label  |   |AOO|   |   |
content|AAA|AAO|   |   |
Cabinet #1 contains two apples and can't be AAO

cabinet| 1 | 2 | 3 | 4 |
label  |AAO|AOO|AAA|   |
content|AAA|AAO|   |   |
Cabinet #1's label can't be AAA

cabinet| 1 | 2 | 3 | 4 |
label  |AAO|AOO|AAA|OOO|
content|AAA|AAO|   |   |

cabinet| 1 | 2 | 3 | 4 |
label  |AAO|AOO|AAA|OOO|
content|AAA|AAO|OOO|   |
Cabinet #4 can't be OOO

cabinet| 1 | 2 | 3 | 4 |
label  |AAO|AOO|AAA|OOO|
content|AAA|AAO|OOO|AOO|

Third person said "no" and we figured out the cabinet's contents with 100% certainty

[LongGone]

[choof]

Originally Posted by Reincarnate

What does the fox say

stepmania

[Reach]

One of my favorite ultra hard problems (some of you might have seen it before, since I've posted it before. IIRC Reincarnate was the only one to solve it correctly, so don't you dare post the answer :P ):

What is the maximum number of cells that cannot further be further subdivided, formed by the interpenetration of 3 cubes?

[Zapmeister]

Originally Posted by Evnoir

@Zapmeister: I couldn't solve the puzzle and this post was just more for laughs than for solutions. When I solved for all the potential numbers, there was only one spot (6th row, 8th column) that had 2 possible numbers. All the rest were 3+ so I gave up lol

here's the clever solution.

the puzzle is symmetric on the main diagonal.

if you flip the puzzle on that diagonal then all that happens is the 1 swaps with 9, the 6 swaps with 7, the 3 swaps with 5, and the 248 all stay where they are.

so if you flip the puzzle along the diagonal and then relabel the numbers like that, you get the same puzzle.

because i told you explicitly that the puzzle has a unique solution, the solution grid must have the property that if you flip it and relabel the numbers, you get the same solution.

in particular the numbers on the diagonal can't change, so they have to be 2s, 4s or 8s. once you know this the puzzle solves with singles and pairs.

bruteforcing will get you nowhere because SE rates it 11.3 which makes it significantly harder than the puzzle in awein999's thread, and there is no solution to this similar to how i solved that puzzle, because this one doesn't have any exotic patterns up to five numbers.


edit: the answer to my other puzzle was this

[Snowcrafta]

12^2 (4+16/6x4) / 12-(4-2) x 0.5

[MinaciousGrace]

Originally Posted by MinaciousGrace

y do they call it circular logic

becuz that's what it's called

[rushyrulz]

Easy:

Flip this triangle upside down, but you may only touch 3 discs.

Medium:
What is the largest number you cannot make by adding multiples of 6, 9, and 20?

Medium-Hard:
What is the next number in the sequence?

1
11
21
1211
111221
312211

[leonid]

Easy

   A
  O O
 O O O
B O O C

A O O B
 O O O
  O O
   C

Medium

43 cannot be made from 6, 9, 20

44 = 20 + 6x4
45 = 9x5
46 = 20x2 + 6
47 = 20 + 9x3
48 = 6x8
49 = 20x2 + 9

50 and higher can be made with 44~49 + 6xN

So the largest number that cannot be made is 43

Medium-Hard

1
11 One 1
21 Two 1's
1211 One 2 and One 1
111221 One 1, One 2, and Two 1's
312211 Three 1's, Two 2's, and One 1
13112221 One 3, One 1, Two 2's, and Two 1's

I already knew about this sequence so I didn't really solve it

[leonid]

Here's one of my favorite math puzzles:

Me: Here are integers A and B, both greater than 1. Bob is given AxB and Sam is given A+B.
Me: Bob, do you know what A and B are?
Bob: I don't know.
Me: Bob doesn't know. What about you, Sam?
Sam: I don't know, either.
Bob: Ah! I know what they are.
Sam: I as well know what they are.

What are A and B?

EDIT: Let's just say A<=B to prevent confusion

[___________]

A census taker approaches a house and asks the woman who answers the door "How many children do you have, and what are their ages?"

Woman: "I have three children, the product of their ages are 36, the sum of their ages are equal to the address of the house next door."

The census taker walks next door, comes back and says "I need more information."

The woman replies "I have to go, my oldest child is sleeping upstairs."

Census taker: "Thank you, I now have everything I need."

What are the ages of each of the three children?

[VisD]

2,2,9

[igotrhythm]

Originally Posted by leonid

Here's one of my favorite math puzzles:

Me: Here are integers A and B, both greater than 1. Bob is given AxB and Sam is given A+B.
Me: Bob, do you know what A and B are?
Bob: I don't know.
Me: Bob doesn't know. What about you, Sam?
Sam: I don't know, either.
Bob: Ah! I know what they are.
Sam: I as well know what they are.

What are A and B?

EDIT: Let's just say A<=B to prevent confusion

If Bob has AB, then AB must be a number that has more than one way of being factored into two numbers each greater than 1. Thus, the smallest number Bob could have gotten is 12, which has factorization of 3*4 and 2*6. Thus, he can't give an answer. Since Bob can't answer, Sam can't answer either because his number likewise has more than one breakdown of A+B from the given integer...only this time, Sam's number can be no smaller than 6.

Bob now realizes that there is a second way of breaking down Sam's given number. The rest of my answer depends on whether Bob and Sam know each other's numbers.

[leonid]

They don't know each other's numbers, otherwise their conversation is meaningless as they have the same information

[L.B.D.D]

does it look like i know what a jpeg is

[___________]

Originally Posted by leonid

Here's one of my favorite math puzzles:

Me: Here are integers A and B, both greater than 1. Bob is given AxB and Sam is given A+B.
Me: Bob, do you know what A and B are?
Bob: I don't know.

As igotrhythm said, this means AB can be broken down in more than one (proper) way.

Originally Posted by leonid

Me: Bob doesn't know. What about you, Sam?
Sam: I don't know, either.

Bob's combinations of possible A's and B's also both give numbers that can be broken down in more than one (proper) way.

Originally Posted by leonid

Bob: Ah! I know what they are.
Sam: I as well know what they are.

What are A and B?

A+B can't be too high nor too low. A little bit of guess-and-checking gives A = 2 and B = 6. We see that this pair of numbers satisfies all requirements.

[igotrhythm]

The problem with Underscoreman's solution is that those aren't the only numbers that satisfy the requirements. A=3 and B=4 would do it as well.

[___________]

False. 2*5 = 10 can only be broken down in one way (other than 1*10)