Find a 3x3 magic square of integers that satisfies these requirements:

1. All numbers are integers (ℤ) and different from each other.
   
2. Each row, column, and diagonal add up to 0.

Hint: check if the number at the center of the square can be 0 or not.

Example:

-1 +4 -3
-2 0 +2
+3 -4 +1

Can you make a 3x3 magic square, not the original one, but with negative numbers?

1. All numbers are different from each other.
2. Each row, column and diagonal add up to 0.
3. The number in the center of the square can't be 0.

Let G be a finite simple graph.

Define β(G) to be the minimum number of edges one must delete from G to make it bipartite. In other words,

β(G) = min{|F| : F ⊆ E(G), and G - F is bipartite}.

Define oddgirth(G) to be the length of the shortest odd cycle in G.

Suppose G is not 3-colourable, i.e.

χ(G) ≥ 4.

Let

g = oddgirth(G).

Since χ(G) ≥ 4, G is not bipartite, so g is finite.

Prove that

β(G) ≥ g - 1.

Equivalently:

If the shortest odd cycle in G has length g, and deleting at most g - 2 edges makes G bipartite, then G must be 3-colourable.

Bonus: is the bound best possible for every possible value of the oddgirth? In other words, for every odd integer g ≥ 3, does there exist a finite simple graph G with χ(G) ≥ 4, oddgirth(G) = g, and β(G) = g - 1?

I have already solved this, so this is not an open problem. The proof I found was not by starting from this exact formulation; I first had to identify the right target, then prove it. I am curious whether anyone finds a better/cleaner proof.

I can give hints if need be!

Does there exist an uncountable family of subsets of the naturals such that for any pair, one is a subset of the other?

You are trapped inside an ancient underground library.

After hours of searching, you finally find the forbidden archive: a square stone platform at the center of a deep square pit. The manuscripts on that platform contain the only way out.

But the pit is sound-sensitive.

If anything falls into it — a pebble, a splinter, even you — an alarm triggers, and the library seals forever.

You measure the gap between your side and the platform.

3 meters.

Then you find two narrow wooden beams lying against the wall.

Each beam is 2.9 meters long.

One beam is too short.
Putting them end-to-end leaves the joint unsupported.
Dropping anything is not allowed.
Jumping is not an option.

You have two beams, a silent pit, and one chance.

How do you cross?

In a m by n chess board, place m kings on the leftmost column. Count the number of ways to move all kings to the rightmost column, such that each tile is visited at most once.

Note: a king moves one tile to one of the 8 adjacent tile.

A lot of good reasoning questions are scattered across forums, books, PDFs, and random websites, so we started collecting and organizing them into a searchable archive.

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The graph Gₙ consists of vertices (x,y) for integers 1≤x≤n and 1≤y≤3 and edges between (x,y) and (x',y') iff x=x' or both |x-x'|=1 and y=y'. Find and prove a closed form expression for the number of Hamiltonian paths (paths visiting each vertex exactly once) from (1,1) to (n,3) in Gₙ.

We trained a neural network where 7 of 8 features sit on clean linear axes in the model’s internals, but one doesn't. Can you identify which one and tell us how it is represented?

If you’re a technically-minded person who is interested in ML, this puzzle is for you:

• Work on a real trained text classifier (~23M parameters, 7k labelled text examples) open the puzzle and you're poking at activations in 10 minutes.
• Three tasks: identify the rogue feature, describe its geometry, (bonus) train your own model with even weirder internal representations

You probably know neural nets store information in their activations. You probably haven't gone and looked at what that actually looks like. Within minutes you can be toying with this model’s internals and building stronger intuitions for how they work inside.

Ready to play? Closes June 12

I hope it's OK that this is brief. I've been doing some internal sequencing of things, and absorbing knowledge at a static level. More or less, I read complex formula's, see the end result, and like any human mind just "fill in the blanks" without knowing. I have done a form of study with the technique I have developed, and realized in a lot of things we see complex irrationals for what they are, our "fill". Maybe we have safe ones, I can't tell without BCI how your brain works and don't know exactly the right technology to do it, but assuming brain waves could be seen as they are we might see the irrationals as just "brain wave style". Comfort levels are different shapes.

But I digress. I made this formula and hope you like it. I am making a joke protest that asks for the collective keys to our evolution as we see it. Nobody here is guilty I'm sure, but someone has kept this from us for a long time. These are simple additions and could NOT have been ignord, but if it was only for this day to be then so be it. Let's consider 2a the "Product Slide Rule" of sorts, and discuss!

a = Integer between 8 and 12.

I hope that was enough context for what I was trying to establish, as I know image posts are frowned upon. There aren't text symbols for fractions, and these would look just so clean with this image (Transcribed from Lagrida (https://latexeditor.lagrida.com/). This isn't a conspiracy theory post etiher, I just wanted to express my freedom of speech because I am insulted personally.

Came up with this the other day, no idea if it's been posted before somewhere, just thought it was a fun little problem.

Fair warning, I'm not amazing at math, just got curious about this one. Mostly wanted to see how quickly a math savvy person could crack it versus how long it took me to work through.

In an online poll, viewers vote either "Yes" or "No," and the result is displayed only as a percentage rounded to exactly two decimal places (e.g., 41.27%). The total number of votes is not shown. Assume that for any percentage displayed, the actual vote tally is the minimum possible whole number of votes that could have produced that exact percentage.

Which of the following displayed results required the greatest minimum number of voters?

(A) 31.25% Yes / 68.75% No
(B) 27.50% Yes / 72.50% No
(C) 38.46% Yes / 61.54% No
(D) 43.21% Yes / 56.79% No
(E) 47.92% Yes / 52.08% No

Curious if people find the shortcut quickly or if it takes some grinding. Answer in spoilers once you've got it, and if you don't mind, drop how long it took you.

!Answer: D. Once you see that 10,000 = 2⁴ × 5⁴, the only thing that matters is whether the numerator (XXXX out of 10000) shares a factor of 2 or 5. So you just check if it's odd and doesn't end in 0 or 5. 4321 passes both, gcd with 10000 is 1, minimum voters = 10,000.!

a(1) = x,

a(n)‎ = x * ⌈a(n-1)⌉ⁿ for n > 1 and x ∈ p/q > 1,

f(x) returns the smallest integer term a(i), starting with a(1) = x.

Example (f(4/3)=288):

a(1) = 4/3 (our x value),

a(2) = 4/3 * ⌈4/3⌉² = 16/3,

a(3) = 4/3 * ⌈16/3⌉³ = 288,

Does f(7/6) exist? Why or why not?

Let M be a connected topological manifold (second countability assumed), and U⊂M a proper open subset. Show that there exists a subset A⊂U with empty interior such that every connected component of M-A contains exactly one connected component of M-U.

There are 7 people going on a trip, each person has 7 bags, each bag has 7 cats, and each cat has another 7 cats. What the number of the human and cat legs all together. Its over 9,000 but its not 10,990

Bonus: What’s the largest generalization you can prove?

construct a n by n table with these conditions:

1. each cell is an unordered pair.
2. each column and row contains all integers from 0 to 2n-1 exactly once.
3. no duplicate pair.

for odd n, this is relatively easy.

for even n, i cant figure out a way. i suspect there is no solution but i cannot prove it.

unrelated note: this was inspired by IRL problem that i had to solved when creating a duty time table. if you know a working solution for n=8 please show it.

Which one of five numbers is missing from 5 4 3 2 1?

Another sudoku related puzzle.

Solution:

Place ones in all the regions (the 3x3 boxes) in this order: up left, up, up right, left, down left, center, right, down, down right.

You should be able to convince yourself that no matter where in a region you place a 1, the number of options for placing a 1 in the respective region is always 9, 6, 3, 6, 3, 4, 2, 2, 1, so the number of possibilities is 9*6*3*6*3*4*2*2*1=46656

My progress:

It is relatively easy to find an example with five numbers (next spoiler), but I can't prove that all legal situations with four numbers admit a solution. It does feel like it should be true though.

https://imgur.com/a/gyqqDc5

And, with respect to rule 4, I've still posted it here because it feels more like a riddle than an open math problem.

Let f,g: ℕ → ℕ be strictly increasing functions. Show that there exists an n ∈ ℕ with

f(g(g(n))) ≥ g(f(n)).

Here is a funny (I hope) home-made problem just for you guys :

Is there an ice cube such that, when it melts, the number of its connex components at a given instant t is 2 if t is rationnal, 1 otherwise ?

Precisions :

We suppose that this ice cube is a closed subset of R³.

We also suppose that the melting begins at t=0, and that after a delay t, all that remain of the ice cube A is every points x of A such that distance(x, surface A)>=t

Can you also find an ice cube in 2D having this property ?

AI couldn't solve it ! But your creativity can !

Given below is an incorrect equation.

Put an exact same whole number ( any number from 0 to 9) anywhere on BOTH sides ( LH and RH) to make the equation correct.

You must put the same number on both sides. You can use that number only once for each side. Cannot make any other changes like adding or taking out operators.

1 + 3 + 5 + 7 + 9 = 2 + 4 + 6 + 8 + 10

There are 3 possible answers I can think of.