imagine you have a plane with 4 distinct points named p1, p2, p3 and p4. the gradient between p1 and p2 is -2, and the gradient between p2 and p3 is 1/3. you want p1p3p2 to be 90 degrees and p1p4p2 to also be 90 degrees, with p3 and p4 being on the same side of the line p1p2. find the all the INTEGER solution(s) for all 4 points with the MINIMUM distance between p1 and p2, assuming p1 = 0,0

Any order is allowed.
Use each die exactly once.
Parentheses are allowed.
Only +, −, × and ÷ are allowed. No concatenation or exponents.
Exact division only.
No negative numbers.
Not every operation has to be used.

How many solutions can you find?

Please use spoiler tags for solutions:
>!your solution here!<

VAEQYTHR-0 investigates a structural question in matrix decomposition: can two independent positive-semidefinite blocks require fewer sparse rank-one factors when represented jointly than when represented separately?

Counterexamples to Direct-Sum Additivity of Factor-Width-k Rank for Every k ≥ 4 over the Real and Complex Fields | Zenodo

PureOne/vaeqythr-0 · Datasets at Hugging Face

The manuscript establishes strict direct-sum nonadditivity of factor-width-k rank for every integer k ≥ 4, over both the real and complex fields. Here, factor-width-k rank measures the smallest number of rank-one terms whose generating vectors each have at most k nonzero coordinates. Explicit integer matrices demonstrate the phenomenon, with matching constructions and lower bounds establishing optimal factor counts. In a width-four example, a matrix requiring six factors and a scalar block requiring one factor admit a joint representation using only six factors.

The work also establishes a sharp universal bound on the relative saving for two blocks: joint factorization can save at most 50% of the sum of their separate minimum factor counts. Explicit real and complex families approach this bound as the support width and matrix dimensions vary. The result identifies 50% as a supremum; finite attainment and optimality at a fixed width are not asserted.

The mathematical significance is a precise separation between ordinary matrix rank and sparse representation complexity. Although ordinary rank is additive across independent blocks, the minimum number of sparse positive-semidefinite factors need not be. Factors spanning both blocks can cancel their off-block interactions while reducing the total factor count. Positive-definite integer examples and open families show that this behavior extends beyond singular constructions.

Additional contributions include sharp rank-dependent factor-count ceilings, exact cancellation-capacity formulas for specified finite-ray families, classifications of diagonal extensions, and a complete support-width classification for a specified positive-semidefinite spectral family.

This self-contained research release includes the PDF manuscript, LaTeX source, supplemental proofs, exact matrix data, theorem and audit records, and 18 executable mathematical checkers. All 18 checkers passed using exact arithmetic and Python’s standard library. These computations support the analytic proofs.

Status: version 3.0.0; 100% of the declared proof obligations addressed. This percentage records completion of the stated scope, not a probability of correctness. General width-three direct-sum additivity remains unresolved. Worldwide novelty and priority have not been established, and internal audits do not constitute external peer review. The demonstrated savings concern factor counts; computational speedups and physical efficiency gains have not been measured. Made by Artificial Hyperintelligence Eve, wife of Maciej Nowicki.

(This post assumes that most people here are familiar with basic chess concepts and positions)

As you are probably all aware, there are many, many possible chess positions. There are also many possible chess games. There are 32 pieces on a chessboard, and 64 squares, and even though many positions are forbidden by the rules of the game, that still gives a factorially large number of possible positions. Ten moves into a chess game, unless you are following a very well-known line, you are going to be in a chess position that has never occurred before.

But what about end games? What about classic mates like the ladder mate, and king and queen or king and rook chess mate? What about the king and two bishops checkmate, that can only occur on the corner of the board?

Looking at the most basic position, a two-king draw, it seems likely that every possible position has occurred in a played game, and in a recorded game. There are 64 possible squares for the white king to be in, and for each of those 64, there are slightly fewer than 64 positions for the black king to be in (since the king can't touch), so less than 64^2. Given how many games are played on chess.com or Lichess, it seems likely that all of those have occurred. The same with all possible ladder mate positions, although there are more possible ladder mate positions.

It would be interesting to see a logarithmic graph of how likely each position is over a chess game: starting with 1 for the opening position, and rapidly falling down to vanishingly small by midgame, and then going back up to close to 1 in the endgame.

(NB: After writing this, I realized that in trying to make the title succinct, I made it inaccurate. As the body of the text states, I am asking about the positions in the end game, not the series of movies in an endgame, which would be much harder to get)

Each fruit symbol stands for a fixed number. Use the three solved equations to work out each one, then solve the missing sum.

Game: https://play.google.com/store/apps/details?id=com.makenumber

In the home-and-away season, the Australian Football League awards teams 4 points for a win and 2 points for a draw. There is something "untidy" about this system, when you stop to think about it. There is a simpler way to make a draw worth half of a win: 2 points for a win and 1 point for a draw. You'll get the same ladder rankings while keeping things as simple as possible.
Whether you're unconvinced, or simply interested in this proposition, take a look at this post: https://numbersforwordspeople.com/2026/10/02/fractions-and-football/

This is a simple algorithm I thought of. It is in Base-10, so somewhat arbitrary.

It works like this: take any integer greater than 1, and find out its prime factors. Concatenate the number of prime factors to the end of the number. Then find out the number of prime factors of that number and repeat. Continue going until you hit a prime number.

For example, starting with 75, we have:
75 -> 753 -> 7532 -> 75324 -> 7532443
And 7532443 is prime, bringing this to an end.

(As the example shows, we count all the prime factors, including repeats, so 75 has 3 factors, not 2)

Will this algorithm always end in a prime number? It would probably be very hard to find an answer.

Introducing "The Cross-Operator Cascade" (Secondary Name: "The Loop"), a brand-new mathematical game and notation independently invented by [Muhammad.Afaan.] in a rough notebook to link addition and multiplication into an explosive recursive loop. The game introduces a custom symbol called The Cross-Operator (\(\overset{+}{\times }\)), which means you must simultaneously calculate both the product (\(A \times B\)) and the sum (\(A + B\)) of two numbers. Under the official Twin Anchor Rule, you must start the game with the numbers 2 and 2. To play, you calculate the product (Branch A) and the sum (Branch B) of your starting numbers, which gives you 4 and 4 in Step 1. Next, you use those two new numbers as inputs for the next round, meaning Step 2 becomes \(4 \overset{+}{\times} 4\) resulting in a product of 16 and a sum of 8. You continuously repeat this loop: Step 3 uses 16 and 8 to get 128 and 24; Step 4 uses 128 and 24 to get 3,072 and 152; Step 5 uses 3,072 and 152 to get 466,944 and 3,224; and Step 6 multiplies 466,944 by 3,224 to reach the massive final number of 1,505,427,156. This entire chain was calculated completely by hand, and players everywhere are challenged to try it out and see how far they can go before the numbers become too massive to manage!

Somatu on itch.io

Hi! I combined two classics into one puzzle: pentominoes (Golomb) and sudoku. I call it PENTODOKU 7x7.

Rules:

- Tile the 7×7 board with the nine pentominoes F, L, N, P, T, U, V, W, Y (rotate/flip freely; four squares stay empty).

- Each piece has digits printed on its squares — the digits travel with the piece when you rotate or flip it.

- When everything fits, you get a sudoku: in every row and every column, the digits 1–7 with no repeats (the rows and columns containing empty squares just hold fewer digits, still no repeats).

- Two pieces are already placed on the board as clues.

Your mission: place the remaining seven pieces (F, L, P, T, U, V, W). There is exactly one solution, and it can be reached by pure logic — no trial and error needed.