Playable link: https://mathfights.app/

I created MathFights, a platform designed to make mathematics more competitive, starting with its first game: Mental Math.

The idea is simple: if you think you're faster at mental math than everyone else, here's your chance to prove it!

In Ranked matches, two players compete in real time, solving mental arithmetic problems as quickly as possible. Both players receive exactly the same questions in the same order, so the competition is fair.

There are three difficulty levels — Standard, Intermediate, and Advanced — and matches last either 1 or 3 minutes.

There are several ways to play:

• Ranked: Compete against other players, climb the leaderboards, and improve your Elo rating.
• Practice: Train solo at your preferred difficulty level.
• Challenge a Friend: Invite friends to private matches, even without an account.
• Ghost Battles: Compete against recorded performances from previous opponents.
• Daily Challenge: Take on a new daily challenge at your preferred difficulty.

My goal is to make mental math more competitive and engaging.

I'd genuinely appreciate any feedback.

21 October is Martin Gardner's birthday, which puzzle fans celebrate as Celebration of Mind Day. Here's a wheel puzzle for it.

Trace a path through the wheel: number, operator, number, operator, … ending on a number. You can jump from any tile to any tile, but each tile can be used only once (or not used at all).

One rule to keep in mind: the path is calculated left to right, like a basic pocket calculator. So 3 → + → 6 → × → 7 evaluates as (3 + 6) × 7 = 63.

Targets to solve: 25, 34, 112, 144, 175. The video solves 25 with path 3 → × → 9 → − → 2. Find a path for each of the other four.

Bonus: every one of these numbers is connected to Gardner or to the puzzles he loved — two are about magic squares, one about Fibonacci, and two about Gardner himself. Try matching them before you peek:

- 25:  the years Gardner wrote the "Mathematical Games" column in Scientific American, 1956–1981

- 34:  the magic constant of a 4×4 magic square — every row, column and diagonal adds up to 34

- 112:  Gardner's age today: he was born on 21 October 1914

- 144:  12 × 12, and the only Fibonacci number above 1 that is a perfect square

- 175:  the magic constant of a 7×7 magic square

Please put answers in spoiler tags:  like this

For more puzzles like this, check out CalcWheel game.

Each diamond follows the same rule linking its top, left, right and bottom numbers. Find the rule and work out the ?. Spoiler-tag your guess: answer 🏆 NEW: LEADERBOARD! Think you're good at this? Rule Hunter now has a daily, weekly and all-time leaderboard. Sign in, pick any nickname you like (only your nickname is shown publicly) and see if you can reach #1. Who's taking the top spot? 👉 Rule Hunter: https://play.google.com/store/apps/details?id=com.makenumber

$$f(n) = n^3 + 54n^2 - 2917n + 20021$$ ( 40 primes)

other ones that i found

• $f(n) = n^3 + 57n^2 - 2806n + 17159$ $\rightarrow$ 39 primes
• $f(n) = n^3 + 60n^2 - 2689n + 14411$ $\rightarrow$ 38 primes
• $f(n) = n^3 + 63n^2 - 2566n + 11783$ $\rightarrow$ 37 primes
• $f(n) = n^3 + 66n^2 - 2437n + 9281$ $\rightarrow$ 36 primes
• $f(n) = n^3 + 69n^2 - 2302n + 6911$ $\rightarrow$ 35 primes
• $f(n) = n^3 + 72n^2 - 2161n + 4679$ $\rightarrow$ 34 primes
• $f(n) = n^3 + 120n^2 - 2893n + 2131$ $\rightarrow$ 33 primes
• $f(n) = n^3 + 75n^2 - 2014n + 2591$ $\rightarrow$ 33 primes
• $f(n) = n^3 + 4n^2 - 1081n + 12289$ $\rightarrow$ 33 primes
• $f(n) = n^3 + 78n^2 - 1861n + 653$ $\rightarrow$ 32 primes
• $f(n) = n^3 + 7n^2 - 1070n + 11213$ $\rightarrow$ 32 primes
• $f(n) = n^3 + 125n^2 - 4412n + 27823$ $\rightarrow$ 31 primes
• $f(n) = n^3 + 54n^2 - 2329n + 20507$ $\rightarrow$ 31 primes
• $f(n) = n^3 + 10n^2 - 1053n + 10151$ $\rightarrow$ 31 primes
• $f(n) = n^3 - 20n^2 - 561n + 10223$ $\rightarrow$ 31 primes
• $f(n) = n^3 - 25n^2 - 292n + 4877$ $\rightarrow$ 31 primes
• $f(n) = n^3 + 128n^2 - 4159n + 23537$ $\rightarrow$ 30 primes
• $f(n) = n^3 + 101n^2 - 3080n + 13537$ $\rightarrow$ 30 primes
• $f(n) = n^3 + 71n^2 - 2690n + 14197$ $\rightarrow$ 30 primes
• $f(n) = n^3 + 57n^2 - 2218n + 18233$ $\rightarrow$ 30 primes
• $f(n) = n^3 + 20n^2 - 2099n + 23581$ $\rightarrow$ 30 primes
• $f(n) = n^3 + 23n^2 - 1988n + 24223$ $\rightarrow$ 30 primes
• $f(n) = n^3 + 45n^2 - 1936n + 7307$ $\rightarrow$ 30 primes
• $f(n) = n^3 + 13n^2 - 1030n + 9109$ $\rightarrow$ 30 primes
• $f(n) = n^3 - 17n^2 - 598n + 9643$ $\rightarrow$ 30 primes
• $f(n) = n^3 - 22n^2 - 339n + 4561$ $\rightarrow$ 30 primes
• $f(n) = n^3 - 28n^2 - 119n + 4219$ $\rightarrow$ 30 primes
• $f(n) = n^3 - 37n^2 + 162n + 3517$ $\rightarrow$ 30 primes
• $f(n) = n^3 - 62n^2 + 781n + 227$ $\rightarrow$ 30 primes
• $f(n) = n^3 - 188n^2 + 2161n + 443$ $\rightarrow$ 30 primes

Hey everyone! We created a daily math grid game called QuizSum where you use arithmetic tiles to reach a daily target number. Speed Mode gives you 3 hidden paths to find against the clock, while Pro Mode offers dozens of unique solutions.

Today's Speed Target is 250. Give it a spin and let me know your fastest time!

🔗 
www.quizsum.com

Hey everyone! We created a daily math grid game called QuizSum where you use arithmetic tiles to reach a daily target number. Speed Mode gives you 3 hidden paths to find against the clock, while Pro Mode offers dozens of unique solutions.

Today's Speed Target is 250. Give it a spin and let me know your fastest time!

🔗 
www.quizsum.com

Each pair of numbers produces the number below it by the same hidden rule. Solve for the missing number from the pattern shown by the other pairs. 🏆 NEW: LEADERBOARD! Think you're good at this? Rule Hunter now has a daily, weekly and all-time leaderboard. Sign in, pick any nickname you like (only your nickname is shown publicly) and see if you can reach #1. Who's taking the top spot? 👉 Rule Hunter: https://play.google.com/store/apps/details?id=com.makenumber

シグマの公式から秒でわかるけどなんかめっちゃ美しくね？

Is it just me or when you sum up integers and multiply it by twice the interval divided by the last number you'd get the next integer? For example 1+2+3+4=10. Multiply by 2 (which is twice its interval of 1) =20. Divide by 4 (the last integer) =5 (the next integer).

Thomas Bradwardine in the 14th century, and later deeply formalized by the famous astronomer and mathematician Johannes Kepler in the 17th century. The universal formula you interacted with—{Sum} = (n-2k) \\times 180—is a standard theorem taught

While the original proof is profoundly complex I tried to make it as easy as possible we start by forming a star shaped polygon of 'n' number of tips ( remember the star should contain sharp triangle rather than snow flake shaped star) we would than cutting those tips triangle all of them and now we have n number of triangles we would place those triangles in a rectangle which form n\*2 triangles. We would than let one angles tip be called x all of the tips are congruent to the measure of x because of interior alternate angles thus making the alternative tip of the one angle from each triangle be congruent

We know that sum of all angles in rectangle is 360..( 90°\*4)

Because the n number of triangle take exactly half area of triangle the sum of all tips would be degree of rectangle/2 which would be 180

Small not it doesn't matter whether n is odd or even because by this \*2 rule it would always be a even number

E+E=E

O+O= E

Hope this helped anyone

if we have a 4x4 grid of squares in which we can place letters of the English alphabet, one in each square then find the maximum number of possible 4 letter words formed if repeating letters is allowed. for example.

A F A T

B E S H

M R X E

G D F E

rule to follow: if you choose a letter tile for the starting letter of your word you must choose the tile adjacent to it ( diagonals count) for the second letter of your word and continue until u get a 4 letter word. for example, in the example above if you choose F as the first starting letter u must select any 5 of the tiles surrounding it as the next letter of your 4 letter word. by choosing F then A then T and then S , you get FATS so like this you must form 4 letter words and have to find the maximum number of 4 letter words possible if repeating letters is allowed( as shown, A is repeated in example and so is E)

Hey r/CasualMath!

I’ve been creating a daily browser puzzle called **QuizSum** — it’s a multi-solution mental math challenge inspired by Countdown and Wordle.

🎯 **The Goal:**
Combine grid numbers using +, −, ×, and ÷ to land exactly on the Target Number.

⚡ **Speed Mode:** Find 3 distinct valid equations against a stopwatch.
🏆 **Pro Mode:** Find ALL hidden arithmetic paths in the grid.

It’s 100% free, runs right in your browser with no sign-up or app download, and has automated path checking.

Give today's puzzle a try at **www.quizsum.com\*\* — would love to hear your thoughts or see what paths you find!

Every input should output a single value, so how does ts work?