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31 votes
Accepted

Professor Halfbrain and the sum of the digits of all divisors

Professor Halfbrain's theorem is Proof
hexomino's user avatar
  • 136k
27 votes
Accepted

Averaging numbers on the blackboard

First choose $2014$ and $2016$. Average = $2015$. Now take the $2015$s. Their average is $2015$. Now choose $2015$ and $2013$. Average = $2014$. Choose $2014$ and $2012$. Average = $2013$. Note ...
iamwhoiam's user avatar
  • 1,348
26 votes

Which is larger? $\sqrt{15} - \sqrt{7} + \sqrt{5} + \sqrt{2}$ versus $5$

The answer is I imagine the line of reasoning the author wants is as follows:
hexomino's user avatar
  • 136k
22 votes

Is it possible that the last piece the ant has eaten is the central one?

I will give you a hint rather than an answer (because I think you will enjoy this more by solving it yourself): I would encourage others not to post an outright solution, at least for a few days; ...
Gareth McCaughan's user avatar
19 votes
Accepted

Amnesiac in a ring shaped palace

N = 2 By referencing the wall colors, you should be able to deduce which direction you were going (CW/CCW or R/L) when entering your current room. I used that assumption to come up with a state ...
ajee's user avatar
  • 356
19 votes
Accepted

Exterminating blobs on a grid

Given an arrangement of blobs, how can you determine whether it is possible to exterminate them all? What strategy can you use to succeed when possible? Warning: what follows is a constructive but ...
noedne's user avatar
  • 15.4k
18 votes

Amnesiac in a ring shaped palace

Edit: A lot of credit is due to @ffao for devising a better way to deal with the case where there is just one room and reducing the solution by one. (Subsequently, @Lawrence has managed to do even ...
hexomino's user avatar
  • 136k
16 votes

Will you be the first to get free?

There is a simple solution. Because
Florian F's user avatar
  • 30.6k
15 votes

Amnesiac in a ring shaped palace

N=5 This builds on @ffao's on-on pattern and is a slight optimisation over @hexomino's 7-step 6-step solution. I'm assuming that in a 1-room scenario, stepping out of that room will lead to the ...
Lawrence's user avatar
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15 votes
Accepted

Consecutive Towers of Hanoi

Here is a revised solution, for... ...which  (again) seems like the maximum to me.  has been verified by Molhan as being maximal.   Trivial steps have been condensed. These ...
humn's user avatar
  • 21.9k
15 votes
Accepted

Swapping registers in an old calculator

Task 1 was answered. Task 2:
Anders Kaseorg's user avatar
13 votes
Accepted

Sliding balls on a 5x5 grid

This is definitely a bit more difficult that the ones before, but despite OP's warnings in the comment section, there's no need for any kind of brute force. (There's an answer with a solution already, ...
Bass's user avatar
  • 77.6k
12 votes

Averaging numbers on the blackboard

Reasoning: After this we... ...going all the way back to...
MichaelK's user avatar
  • 709
11 votes

Which is larger? $\sqrt{15} - \sqrt{7} + \sqrt{5} + \sqrt{2}$ versus $5$

Here's another approach, which starts from smaller integer residuals than hexomino's solution. and Finally, bringing these lines of argument togegther, I admit it isn't a single chain of inferences....
Rosie F's user avatar
  • 8,640
10 votes

Sliding balls on a 4x4 grid version 2

@user already posted a solution, mine differs slightly at the end: (spaces inserted at the points where I was mentally switching to another task; the final 5 moves are where the solution differs from ...
Bass's user avatar
  • 77.6k
9 votes
Accepted

Blackboard problem with polynomial

The smallest possible value of $n$ is Claim: We can get every non-negative integer $n\leq 2016$ on the board. Proof: By induction. We start with $n=0$ on the board. We can get $1$ using Lord of the ...
Julian Rosen's user avatar
  • 14.3k
9 votes
Accepted

Sliding balls on a 4x4 grid version 2

Is this a trick question? (New here) Is the answer just Otherwise,
fool's user avatar
  • 206
8 votes

Amnesiac in a ring shaped palace

[EDIT: hexomino points out in their answer that we can just use a completely off pattern instead of an alternating pattern to eliminate most of the states here, which makes this answer unnecessarily ...
ffao's user avatar
  • 21.8k
8 votes
Accepted

Permuting rows and columns to switch white rooks with black rooks

Method:
greenturtle3141's user avatar
7 votes
Accepted

Aatif averages numbers on the blackboard

The answer is : Explanation : Generalization :
Fabich's user avatar
  • 7,165
7 votes
Accepted

Four indeed is cosmic!

Let the starting number be $n$. Consider the case where $n < 10$. $$ \begin{align} 1 &\to 2 \to 4 \\ 2 &\to 4 \\ 3 &\to 6 \to 12 \to 24 \to 2 \to 4\\ 4 \\ 5 &\to 10 \to 1 \to 2 \to ...
Lawrence's user avatar
  • 7,929
7 votes
Accepted

Is it possible that the last piece the ant has eaten is the central one?

A cube of dimension $3×3×3$ is made of sugar and consists of 27 small cubical sugar pieces arranged in the $3×3×3$ pattern. An ant is eating the sugar in such a way that it starts at one of the ...
Ilmari Karonen's user avatar
7 votes
Accepted

Will you be the first to get free?

Yes. Explicit Grid
Display name's user avatar
  • 2,260
6 votes

Consecutive Towers of Hanoi

I wrote a program to find the answer to this question. I indeed found that 31 was the maximum number of disks for a 6 peg board. I also ran the program with 3, 4, and 5 pegs. Interestingly, the ...
Molhan's user avatar
  • 61
6 votes

Professor Halfbrain and the 52 cards

The optimal solution is: First an example of the algorithm with 10 elements: The algorithm used follows: I will also only amend to the other solutions proof to show that the total number of swaps ...
Ewok's user avatar
  • 61
6 votes

Which is larger? $\sqrt{15} - \sqrt{7} + \sqrt{5} + \sqrt{2}$ versus $5$

Yet another method: And then:
Toby Mak's user avatar
  • 494
6 votes
Accepted

Labyrinth of Teleporters

Without the pebble... But... Using the pebble...
tehtmi's user avatar
  • 3,316
5 votes

Is it possible that the last piece the ant has eaten is the central one?

This can be done with various approaches, I have tried using a very simple logical approach here without any previous knowledge required (only a pencil & paper)
Netham's user avatar
  • 372
5 votes
Accepted

Is it possible (for some configuration of initial 9 flowers) to get all red flowers after finitely many years?

The answer is: Slightly more rigorously, for all the flowers to be red: So at some point: Then: Then: But then: So: Note:
boboquack's user avatar
  • 22k
5 votes

Amnesiac in a ring shaped palace

The other answers have done all the hard work, so all that's left is to read the question again to use all the available information to minimise the highest number to remember. I got it down to When ...
Bass's user avatar
  • 77.6k

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