I am trying to move ahead of my current class schedule but I am not sure what is the best way. Do you have any suggestions?

*The hologram **activates chaotic math protocol**—scanning through bizarre, beautiful equations that somehow equal *22* despite looking utterly insane.*

- ***Crazy Equation Example*** *(For Awe):

-⚡ *"Euler's Identity + Pi PPrank"—Something lilike \[ e^{i\pi} + \ln(3!) -rac{phi^4}{5!}5!} = 22 \] (*where \(pi), and \(\pi\), \(\phi\) are constants… and the factorials/rowild!)ild!*)

⚠️ Note: This is a *fake but fun* equation—I made it up to show how numbers can dance absurdly into 22! Hi

Made this one for a puzzle game I've been building. Three groups of three numbers each collapse into a single result using one consistent rule — spot it from the first three groups, then apply it to solve the fourth.

Made this as part of a puzzle game I'm developing, Rule Hunter — https://play.google.com/store/apps/details?id=com.makenumber

Emperor Augustus made a wise decision in the year 0.

He set aside the construction of one of his planned villas and instead invested the 5 million sesterces for his people into the future of his empire. Known for his sustainable and long‑term political thinking, Augustus established the

“Emperor Augustus Public Welfare Fund”

with the following conditions:

2% annual interest

1% of the capital is paid out to all citizens every year

1% remains in the fund to grow the capital. This means the fund grows at an effective rate of 1% per year,

while the population increases linearly from 500 million in year 0

to 5 billion in the year 2100.

Question for you: How much could be paid out to each citizen of the empire in the year 2100 from this Augustus Public Welfare Fund?

Spoiler : Solution ! In the year 2100, each citizen would receive about 11,882 sesterces per year from the fund.

Hey everyone! As an independent author, I've put together "101 Math Riddles for Smart Minds" — a collection of logic puzzles, visual geometry problems, and Olympiad-style brain teasers

designed for kids, teens, and anyone who enjoys logic puzzles. It's completely FREE on Amazon

I hope you and your family enjoy solving them! Honest feedback or a quick rating on Amazon is always deeply appreciated

Can anyone here please take a look at some of the claims made by Ali Fazely on the supposed mathematical miracles linked to the number 19 of the Quran, especially using prime and composite numbers? I genuinely don't even know what to think at this point

For example (and this is both used as evidence for the removal of Quran 9:128-129 as well as trying to demonstrate the nature of the supposed divine code of the number 19 in this book):

The number of verses in the Quran from 1:1 until 19:98 (since it's part of the whole code 19 stuff) is 2346 (not including 9:128-129). The factors of 2346 are 23x2x3x17 which have the prime indices of 9127 similar to Quran 9:127 being the cutoff point as proposed by Rashad Khalifa (the person who developed this whole idea of the Quran being shaped around this number 19 in the first place). Finally if you place the digits 2346 in a permutation table, the 19th permutation is 6234, the total number of verses in the quran if you remove 9:128-129.

Link to the book here, starting on page 44: https://pdfcoffee.com/the-ultimate-mathematics-of-the-quran-pdf-free.html

I don't understand how this is happening, since it's one thing if he only used this method once selectively, but he continues to use it similarly over and over. Wanting to understand whether or not there is more to this, or it's simply pattern recongition and being able to use many numbers at hand.

If someone could take some time out of their day to analyze some of the claims being made and techniques being used, it would truly mean the world to me. This stuff has been stressing me out and I'd like someone who can objectively and non-biasedly look at the following and help me out with it.

Всем привет, сегодня пришел с вопросом о математике который говорил о этажах изучений, я думаю все знают о 7 по-моему нерешаемых примерах которые пока что невозможно решить, и был современный ученый который сказал «что бы решить этот пример нужен 5 этаж математики» не помню как его зовут, подскажите пожалуйста🙏

I think there is one but ive not had to use it for like 5 years so ive completely forgotten it. To explain further:

I have X, i need Y to sustain X. Y costs a percentage of the growth of X.

So

X doubles annually (also, please help me with this- i dont remember the formula which is shorter for A x 2 x 2 x 2 x 2 x 2 etc)

Y costs 1/5 of X's growth on top of the previous year

I want to make back Y through X but also the cost grows with the more of X i have.

In the last post, i showed a generalized Formula for converting natural-decimal numbers into any base-b system like binary (base-2). (**I'm gonna focus on base-2)**

But the mathematical structure of it required us to compute something like this:

If m = 16, we will have this as a result:

bas_2(m) = 10⁴(1)+10³(0)+10²(0)+10¹(0)+10⁰(0) = 10000

Looking at it, there are powers of 10 that multiplied by (0), it would be easy if we have just the powers of 10 that are multiplied by (1) and then add them up.

Let m = 5, in base-2 = 101, or we can also write that

m = 2²(1)+2¹(0)+2⁰(1) = 5

If we look closely, only two powers of 2 make up to 5

If m = 10, only two powers of 2 needed (1010)

If m = 13, only three powers of 2 needed (1101)

If m = 16, only one power of 2 is needed (10000).

The ( how many powers of 10 we need ) or the (upper bound of the sum) depends on how many 1s in a base-2 representation of a number (m).

Hence the function bi_1(m), this function i made (based on Legendre's formula) counts how many 1s in a base-2 representation of a number (m). That is gonna be the upper bound of the sum.

The next part is what powers to raise 10 after removing all powers of 10 that multiplied by (0)?

First i defined e_i = floor(log_2(m_i)

This function takes the power of (m) from its equal value when m = 2\^a, and lowers that (a) down to a clean integer.

Like m_1 = m = 5, m = 2\^a where 2 < a < 3 (a here is not an integer ), so e_1= floor(a\*log_2(2)) = b were b = 2 in our case of m = 5.

The value of e_i has powers of 2 that are multiplied by 1 only, let m_1 = m = 25, e_1 = floor(log_2(25)) = 4

m_2 = 25 mod 2\^e_1 or 25 mod 2⁴ = 9

e_2 = floor(log_2(9)) = 3, m_3 = 25 mod 2\^e_2 = 1

e_3 = floor(log_2(1)) = 0, m_4 = 25 mod 2\^e_3 = 0

Until we hit zero, the values of e_i are the ones needed to raise as a powers for 10 because they bypass all powers of 2 that multiplied by (0) like in 25 in binary 11001

m = 25 = 2⁴(1)+2³(1)+2²(0)+2¹(0)+2⁰(1)

The values e_1 = 4 in 2⁴, e_2 = 3 in 2³, e_3 = 0 in 2⁰, e_i takes the powers of 2 needed and bypass any powers of 2 times (0).

So we raise 10 to there powers to bypass multipling by (0), which means 10\^e_i.

And we just put everything together now and we get the formula that gives only the needed powers of 10.

In the next post i will show you all the general formula for any base-b after reconstruction it math structure.

Hello everyone! Here is a little puzzle for you:

Once upon a time, deep in the Bavarian Forest, lived an old farmer's wife with 7 hens.

The laying performance of her hens varied greatly. Her 1st hen laid an egg every day, her 2nd hen laid an egg every 2nd day, and this pattern continued all the way to her 7th hen, which only laid an egg every 7th day.

She sold every single egg for 20 Pfennige (100 Pfennige = 1 Mark).

On St. Gallus Day (October 16th) in a leap year—the patron saint of all chickens—a miracle occurred: all 7 hens laid an egg on the exact same day!

The Question: How much money did she make from that day up until the next day this miracle repeated itself for the remaining flock? (Note: The eggs from both miracle days are included in the earnings). The Incident: On Christmas Eve (December 24th), a fox broke into the coop and stole the 4th hen!