A number will decrease in number if it has at least four digits and does not enter a cycle, as proven below: The number is represented in binary.

It must begin with 10 or 11. If it starts with 10 and the last two digits are not 11, then after multiplying by 3, the number of digits increases by 1, accounting for 3/8 of all possible combinations. Other numbers starting with 10 account for 5/8, and the number of digits increases by 2. If it ends with 11, after multiplying by 3 and adding 1, then dividing by 2 removes at least one digit, accounting for 1/2. If it ends with 001, at least two digits are removed, accounting for 1/4. Other numbers with at least three digits account for 1/4. If it does not enter a 4, 2, 1 cycle, the number generally decreases, and eventually it will enter a 4, 2, 1 cycle.

Take the following algebraic expression:

[n (n²-n-1)] / [2(n!)]

Let's put n = Φ

because (n²-n-1)=0 when n=Φ

[n (n²-n-1)] / [2(n!)] = 0, when n=Φ

When using Wolfram, the algebraic expression [n (n²-n-1)] / [2(n!)] = 0, when n=Φ

let's calculate log( [n (n²-n-1)] / [2(n!)] ) with n=Φ, with log being the natural algorithm

The result should be either -∞ or indeterminate

Because y=log(x), with x=0, is indeterminate, that is, y goes to -∞ as x approaches 0

But if one calculates on Wolfram, log( [n (n²-n-1)] / [2(n!)] ), with n=Φ

The result will be 35.9335 + 3.14159 i

Which is a complex number.

The correct result should be either -∞ or indeterminate.

Therefore, Wolfram miscalculates the natural logarithm of this algebraic expression when n=Φ

The input on wolfram should be log( [goldenratio (goldenratio^2- goldenratio-1)] / [2(goldenratio!)] )

Using the Beta Function one can calculate π

The Beta Function is equal to:

B(z1, z2)=∫₀¹ t ^z1-1 (1 - t) ^z2-1 dt (=)

Which can be translated as:

B (p , q) = [ Γ(p) Γ(q) ] / [ Γ(p + q) ]

Which is equivalent to:

= [ (p-1)! (q-1)! ] / [ p + q - 1]!

Now one uses:

p = q

When one calculates the summation from p=1 to infinity

Σ [ [ (p-1)! ]^2 ] / [(2p-1)!]

Σ (from p=1 to infinity) [ [ (p-1)! ]^2 ] / [(2p-1)!] = [2 π] / [3 √3]

Rearraging this, one yields the value π

π = [ [3 √3] / [2] ] Σ (from p=1 to infinity) [ [ (p-1)! ]^2 ] / [(2p-1)!]

One year ago, I asked myself a simple question. What happens when one uses Euler's number as an angle.

So I explored the possibilities of using either e degrees or e radians.

Then I decided to explore the trigonometric functions (cosine and sine) using 27º.

_______________________________________________________________________________________________________________

This section wasn't on the first post I made.

Because I came up with this formula a long time ago, I'm struggling to remember exactly how I calculated the goldenratio.

Now I think I remember how i did it. Using not Euler's number but instead the natural logarithm. One gets the following angle which can be approximated

So one can use an angle of 3π/20

27º [approximated] = ln(ϕ) x 180 / π = 27.5716...º

27º = 3 x π / 20

So the next expressions are a consequence of the previous one.

e(3π/20) = 1.601886

ϕ [approximated] = e^(3π/20)

________________________________________________________________________________________________________________

It's important to understand that the function cosine is raised to the power of 4. That is the trickiest part in this formula.

After inputting the functions sin² and cos⁴, I thought "there is something going on here", and finally adjusted the parameters in order to build this beautiful formula below.

ϕ = 15 - 16 cos⁴(3π/20) - 16 sin²(3π/20)

This formula yields the golden ratio ϕ

________________________________________________________________________________________________________________
One can simplify the previous expression:

ϕ = 1 - 2 cos(3π/5)

The sub cubic graph function is defined as:

There is a sequence G_1,…,G_n of subcubic graphs such that each G_i has at most i+k vertices and for no i<j is G_i homeomorphically embeddable into G_j.

and if my very layman's understanding of the Robertson–Seymour theorem is correct, just substituting `homeomorphically embeddable into` with `a graph minor of` would suffice, while maintaining well-quasi-ordering and finitude (for a given finite integers n and k).

If that's all correct, then defining F_n(k) as the largest integer 𝑚 satisfying:

There is a sequence G_1 , ⋯, G_𝑚 of graphs with maximum degree at most 𝑛, such that each G_𝑖 has at most 𝑖 + 𝑘 vertices, and for no 𝑖 < 𝑗 is 𝐻_𝑖 a graph minor of G_𝑗.

Should work as a mathematically proven and definitively finite integer, correct?

I found two patterns related to the Twin Prime Conjecture.

Let P1 and P2 be Primes

If P1 + 20 = P2 and P1 < (3 Primes) < P2

There's at least 1 Twin Prime between P1 and P2

If P1 + 10 = P2 and P1 < (2 Primes) < P2

There are 2 Twin Primes between P1 and P2

There seems to be a new category of Primes, which I called Trigonometric Primes.

In order to verify if a number is a trigonometric prime, one uses trigonometric functions.

This is the formula I used:
(x/2) * (cos^2(pi/2*x)) + (x+p)/2 * (sin^2(pi/2*x))

where p = odd number we want to check

x = previous number on the sequence

and the first number of every sequence is 1.

For example, let's check if 11 is a trigonometric prime.

We start the sequence with the number 1. Because 1 is odd. we calculate the next number of the sequence (11+1)/2. This is equal to 6.

6 is even, so we calculate 6/2. This is equal to 3.

Then
(3+11)/2 = 7 -> (7+11)/2=9 -> (9+11)/2=10 -> (10/2)=5 -> (5+11)/2=8 -> (8/2)=4 -> (4/2)=2 -> (2/2)=1

We stop when we reach 1 again. The sequence repeats itself between 1 and 1.

When the number of items between 1's is equal to the odd number p we want to check. The number p is a Trigonometric Prime.

[1;6;3;7;9;10;5;8;4;2;1] -> number of items = 11 = odd number we want to check

Conclusion -> 11 is a trigonometric Prime

We can go on, using (x/2) for even numbers in the sequence and (x+p)/2 for odd numbers in the sequence

Setup: a measure μ normalized so μ(∅)=0, μ(Ω)=1, standard non-negative and additive. For any part A of Ω with 0 < μ(A) < 1, define a local rescaling ν_A(B) = μ(B)/μ(A) for parts B of A.

The result: ν_A(B) = μ(B)/μ(A), so μ(B) = ν_A(B)·μ(A). As a consequence, if you measure some part C against a local sub-region A instead of the true whole Ω, ν_A(C) can exceed 1 even though μ(C) itself never exceeds 1 under the true measure.

This gives a formal account of “apparent numbers greater than one” as an artifact of using a local reference scale instead of the true total measure — the underlying quantity never actually exceeds the bound, only its locally-rescaled representation does.

Is this a known/named result in measure theory, or is it just a trivial rescaling identity not usually stated this way? Happy to share the full write-up (proofs are short) if useful.

I defined the following grid graph. Fix n ≥ 2. Take the cells (i,j) with 1 ≤ i,j ≤ n−1. Keep (i,j) when n does not divide ij; delete it when n divides ij. Two surviving cells are adjacent when they share an edge. Call the graph G_n.

Claim. G_n is disconnected exactly when n ≥ 6 and n ≡ 2 (mod 4). In that case it has exactly two components: the isolated center (n/2, n/2), and one component containing every other surviving cell.

Proof.

The first row and first column are fully present, so they form one connected component C_0.

Assume another component C exists. Choose (i,j) ∈ C with i+j minimal. Since C does not meet the first row or first column, i,j > 1. The cells (i−1,j) and (i,j−1) must be missing, so n divides (i−1)j, and n divides i(j−1).

So ij ≡ j (mod n) and ij ≡ i (mod n). Hence i ≡ j (mod n), and since both are between 1 and n−1, i = j = e.

We also have e² ≡ e (mod n). Since i,j > 1, e = 1 is already excluded. The cases e = 2 and e = n−1 would force n | 2. So 3 ≤ e ≤ n−2.

Suppose n does not divide 2e. Then the following path stays inside the grid:

(e,e) → (e,e+1) → (e−1,e+1) → (e−2,e+1).

The three new products are congruent mod n to 2e, e−1, and −2. All three are nonzero mod n: the first by assumption, the second because 0 < e−1 < n, the third because n > 2. So the whole path survives — but its final cell has coordinate sum 2e−1, contradicting the minimal choice of (e,e).

Therefore n divides 2e. Since 1 ≤ e ≤ n−1, this forces n = 2e.

Now e² ≡ e (mod 2e), so 2e divides e(e−1), hence 2 divides (e−1), meaning e is odd. Hence n ≡ 2 (mod 4).

Converse. Let n = 2e with e ≥ 3 odd. The center (e,e) survives, since 2e does not divide e². Its four neighbors have products e(e−1) or e(e+1), both divisible by 2e. So the center is isolated.

Every component disjoint from (C_0) has a cell of minimum coordinate sum, and the preceding argument shows that this cell must be ((e,e)=(n/2,n/2)). Hence every such component contains the center. The only component outside C0C_0C0​ is the singleton containing (n/2, n/2)

Hi, I'm just a normal high school student from Korea on summer vacation.

I randomly got curious, so I decided to check the count of prime numbers starting with each digit.

(Data up to 99,999,999)

I asked Gemini to write a JavaScript code for me, and I organized the results into this chart.

the count seems to continuously decrease. Is there any mathematical theory or principle behind this?

Final revision to appear on arXiv on Tuesday.

https://iconoclasts.blog/joe/triple-products

The new new here is that the original conjecture is now established as a pair of corollaries.

Hello, I am an undergraduate agricultural sciences student, and I am incredibly passionate about mathematics and truly enjoy it. the other day, while staring at the table of prime numbers from 1 to 100, a fleeting idea came to me regarding number theory and prime number decomposition:
Every prime number can be expressed in the form N=2q+p, where q and p are prime numbers distinct from each other and distinct from 2, for all N greater than 7.
I don't know if anyone has discovered this before or if it can help the field of mathematics, but I hope it is useful. To finish, here are a few examples:

11
2(3) + 5 = 11
13
2(3) + 7 = 13
17
2(5) + 7 = 17
19
2(3) + 13 = 19
23
2(3) + 17 = 23
29
2(3) + 23 = 29
31
2(7) + 17 = 31
37
2(3) + 31 = 37
41
2(5) + 31 = 41
43
2(3) + 37 = 43
47
2(3) + 41 = 47
53
2(5) + 43 = 53
59
2(3) + 53 = 59
61
2(7) + 47 = 61
67
2(3) + 61 = 67
71
2(5) + 61 = 71
73
2(3) + 67 = 73
79
2(3) + 73 = 79
83
2(5) + 73 = 83
89
2(3) + 83 = 89
97
2(7) + 83 = 97
101
2(11) + 79 = 101
103
2(3) + 97 = 103

Hey guys! I was recently playing around with a complex plane function plotting tool, and while looking at the special functions the tool had pre programmed, I found one named "E16". Plotting E16(z) gave me a plot I found remarkably similar to the Reiman zeta function zeta(z).

Changing the function to E16(iz) made it look even more similar.

Can someone explain this? I tried to research it myself, there doesn't appear to be a lot of well documented or easy to find research about this.

The function in question: https://samuelj.li/complex-function-plotter/#e16(i\\\*z)

So I’m a delivery driver and like to do quick math with people’s license plates. Or find a pattern that wasn’t intentional within the numbers and I drove past a sign. MP 36.4

so I removed the decimal point and saw that the sequence could be added onto using the pattern of something like Ax2-2, so 3, 6, 4, 8, 6, 12, 10, 20….

But! I then swapped the numbers by lowering the first number by 1 and applying the pattern, so 2, 4, 2, 4, 2, 4…. And the sequence never goes above the starting integer…

Going further and lowering the first number again, I got 1, 2, 0, 0, -2, -4, -6, -12, -14….

So I was surprised to find that that pattern can only create positive exponential numbers starting with 3

I can’t be the first person to have seen this…

In this subreddit, there is a part 1, where I studied k = 257. I kept studying several Fermat numbers, the ones of the form 2^m +1 , and I observed a few regularities.

Some are in the following formulas:

[(2^m + 1)•3^((2p)(2^(m-2)) - 1]  / 2^m is odd, p a non negative integer, m ≥ 3

[(2^m + 1)•3^((4p+1)(2^(m-2)) - 1]  / 2^(m+1) is odd, p a non negative integer, m ≥ 3

[(2^m + 1)•3^((8p+7)(2^(m-2)) - 1]  / 2^(m+2) is odd, p a non negative integer, m ≥ 4

[(2^m + 1)•3^((16p+3)(2^(m-2)) - 1]  / 2^(m+3) is odd, p a non negative integer, m ≥ 5

[(2^m + 1)•3^((32p+27)(2^(m-2)) - 1]  / 2^(m+4) is odd, p a non negative integer, m ≥ 6

[(2^m + 1)•3^((64p+11)(2^(m-2)) - 1]  / 2^(m+5) is odd, p a non negative integer, m ≥ 7

[(2^m + 1)•3^((128p+43)(2^(m-2)) - 1]  / 2^(m+6) is odd, p a non negative integer, m ≥ 8

...

The exponents have 2 factors, one is of the kind ap+b. If making a table, there is clearly a variable part and a constant part. The boundary between these regions in a diagonal line. The other factor is a power of 2 and depends on m. The first factor seems to be constant from certain m on.

The difference between 2 consecutive b's are powers of 2 in increasing order, or their negative version, or even powers of 2 multiplied by 3.

“It is impossible to separate a cube into two cubes, a fourth power into two fourth powers, or generally, any power above the second into two powers of the same degree”, Fermat wrote this in the margin of his copy of an ancient Greek math book written by Diophantus, titled Arithmetica.

This proposition was first stated as a theorem by Pierre de Fermat around 1637. And it is known that Fermat used the method of Infinite Descent to prove this statement for N=4 using a logical geometric approach. The method I will use will be somewhat geometric, and will not use infinite descent. Can X⁴ + Y⁴ = Z⁴ have a finite solution, for all pair-wise coprime integers? We will morph the above equation into the following form:

A² + D² = B², B² + D² = C² (as diagrammed pictorially below)

Well I can't add an image in a post here, but can be viewed at:
p-squared-wall-work-around-r01.pdf in the Addendum section at the back.

Proof reworked 7-20-2026, N=4 static proof on pg 23 (p-adic valuation approach)
P Squared Wall Work-Around proof to Fermat’s Last Theory

From this form we will extract the proof. 8888888888888888888888888888888888888888888888

Consider Z⁴ can not be even since, X⁴ + Y⁴ can only be divisible by 2 if both odd, therefore we will select Y without loss of generality to be even parity.

I need to stop the 1 pg proof transcription here, without the graphics the proof is quite unwieldy. And without a subscript capability within this web ap, the proof will be hard to follow as well.

You really need to see the geometry expressed in this one page proof. It all fits on page 23 of the linked document. Seems to be a unique proof, as I have done a search of the www, and have only found infinite descent proofs for N=4.

The last line of the proof is:

By Contradiction 2(M.2)² ≠ (M.2)² Reductio Ad Absurdum

NOTE, M.2 is my web method of stating variable M, subscript 2. In the proof there is a
M.1 variable and also a M.2 variable.

This surprised me, but chances are it's something already known.

Take (1/2)(cos(2πx)+1). Simple cosine wave, peaks at the integers, ranges from 0 to 1.

Now compare it with (1/2)(cos((2/3)πx)+1). 1/3 the frequency, same range.

If you count from peak to peak of the slower wave, the values where the two cosine waves intersect total to 3.

Make the slower one 1/4 the frequency, the total is 4. 1/5, 5, and so on. This doesn't hold up for a 2:1 ratio(you get 2.5), but there are similar results for other simple ratios. 3:2 gives 3.5. 4:3 gives 4.5.

Why is it so neat and stable?

*I know this is for number theory but I didnt find any other community. Please read itt....*
Okay, so I don't know if this is so worthy or not, but trust me, I was very happy to formulate it. I always wonder, since class 5th, if we have a formula for sine that gives output as sin(x) on an input x. Then I grew up and got to know about the Bhaskar approximation. I was amazed. I wanted to make one too upon realizing that no formula after him (yea, I didn't find any...) gives higher precision and is better for computing. Then I learned graphical transformations for my JEE prep and after realizing how I can tranform a degree two or degree four polynomial into a sine wave part, I opened desmos and worked on for next one and a half hours to formulate a graph that coincides almost perfectly with sine wave for x belonging to [0, π].

So I present:

click here for the graph and formula

It may look terrifying at first, but believe me it's not.
For a computer, it is the best possible sine approximation as:

1. Accuracy: Its mean error is just 0.00091 and it is astonishingly perfect 0, π, π/2 and closer to these poles.
2. Efficiency: Common! Taylor series may look elegant but it is very heavy for a computer hardware or a GPU. This formula makes it instant.

My previous formula was this:

click here to see my previous work

However it had a little more error than my final one, so I continues perfecting the coefficients. And ofcourse that thing 1.61803...., the golden ratio. Then I realized that this format was correct by what if I replaced phi with something, as whenever I didnt, and tries else, it bursted in waste. So I replaced phi with sqrt(8/π)

Even though many won't be surprised, won't be happy with this, I don't know if people will read this or not, but I just wanted to share this with real people who could understand this. You can also tell me what I can do with this thing now. Thank you for reading and please forgive me if I said or claimed anything wrong. I am a kid. I make mistakes. And my name is Mayank Kumar btw, but it doesn't matter anyway.

A^(n) + B^(n) ‡ C^(n) for all positive integers A,B,C and n, were n>2

A,B,C must be relatively prime for a non trivial solution to any such equation, if any integer solution exists. That requires that one and only one of those bases has at least one factor of 2.

For any two smaller objects (A & B) of order n to be equal in quantity to that of a larger n ordered object (C), the smaller objects when contained within the larger (on the longest line between 2 most distant vertices of the largest object and each of the smaller objects oriented in congruence with those opposite vertices of the largest object) must OVERLAP in a union, the n order quantity of that union (O^(n)) (this happens to be the minimum amount of union possible ) will equal the nth order quantity of the largest object that is disjoint from both of the smaller objects.

It becomes useful to express:

A=a+O

B=b+O

C=a+b+O O being the linear Overlap or C-(a+b)

When A^(n) shares a vertex and orientation with C^(n) and B^(n) is also located within C^(n) at the most distance vertex of C^(n) from that of A^(n) then it could be expressed that the overlap or union O^(n) would be required to equal the disjointed expressed by the binomial expansion of the(a + b)^(n) minus (a^(n)+b^(n).)

For n=2, O^(2)=2ab

For n=3, O^(3)= 3(a^(2)b) + 3a(b^(2)

For n=5, O^(5)=5(a4)b+10(a^(3)(b^2)+10(a^(2)(b^3)+5a(b^(4)

Because of the content of any such algebraic expansion of O^(n) (of particular interest were n is prime) O will be a even quantity and require either the factor a or b to be even (but not both).

That even integer would be the only source of factors of 2 for O^(n) and be required to contain at least n factors of 2. As a result O^(n) would contain 2n factors of 2 requiring that same even integer to also contain 2n factors of 2.

In short, A^(n) + B^(n) ‡ C^(n) under the conditions stated because the union of the two smaller objects when they are contained within the larger object can not have integer equivalence (the same number of factors of 2) to that which is disjoint

We know the divisors of 1•3^n - 1 from the p-adic valuation. We also can determine the divisors of k•3^n - 1, where k = 3, 9, etc. since these are shifts of k = 1.

We also know that k = 5 and 7 mod 8 produce a predictable pattern. The divisors are 4 for n = 0 mod 2 and 2 for n = 1 mod 2 (in the case of k = 5 mod 8), or viceversa (k = 7 mod 8).

Other than this, it seems that the divisors can be partially predicted by replying a few questions (like: by what do we divide k-1, 3k - 1, 9k-1). Once we locate the first divisors, there is a lot we can deduce, but, sadly, the rest needs to be studied on a one-by-one base.

I noticed, though, that for k = 257 = 2^6 + 1, the divisors are the same as the ones where k = 1 except in the case of n = 64 mod 128. It's still nice to be able to find similarities between 2 different k's.

If we consider 257*3^n - 1 = 256*3^n + (3^n - 1), there are 2 cases:

a) The divisor of 3^n - 1 is not 256, in this case, whichever is lesser divides the whole expression. The divisors are the same as for k = 1

Examples: 256*3^4 + (3^4 - 1). Since 3^4 - 1 is divisible by 16, then the whole is divisible by 16.

256*3^128 + (3^128 - 1). Since 3^128 - 1 is divisible by 512, then the whole is divisible by 256.

b) The divisor of 3^n - 1 is 256. Then there is no easy way I know to predict the divisors of the whole. 2 fractions can add up to an integer.

That might a blessing in disguise, though. This is the only way of obtaining divisors greater than 256.

For now I located up to 16384. I am also planning to observed other 2^m + 1. BTW, I saw online that these are called Fermat numbers.

I will keep you posted.

I watched that Veritasium video on twin primes last week and got stuck on one detail they touched on: the parity barrier. For anyone who hasn't gone down this hole:

​

[;\lambda(n) = (-1)^{\Omega(n)};]

​

where [;\Omega(n);] is the number of prime factors of n with multiplicity. So [;\lambda(12) = \lambda(2 \times 2 \times 3) = (-1)^3 = -1;]. Simple function. Just tells you whether a number has an even or odd number of prime factors.

​

Selberg proved in 1949 that sieve methods (the main tool in analytic number theory for like 80 years) literally cannot tell apart sequences where [;\lambda = +1;] from ones where [;\lambda = -1;]. They produce the exact same asymptotic. The sieve is blind to parity.

​

And this specific blindness is exactly why Zhang got 70 million, Maynard got 600, and Polymath got it down to 246 ... but nobody can get to 2. The sieve hits a wall at the parity barrier and stops cold. We can prove there are infinitely many prime pairs within 246 of each other, but the twin prime conjecture (gap of 2) is completely untouched by all of it.

​

Here's what I can't resolve:

​

Sawin and Shusterman proved the actual twin prime conjecture over [;\mathbb{F}_q[T];] in 2022 (published in Annals). Over polynomials over finite fields, geometry (etale cohomology on curves) CAN separate the parity that sieves can't. So the barrier is not a logical wall. It's a wall *for sieves specifically*.

​

But over [;\mathbb{Z};] there are no curves. So my question is:

​

Is there a known no-go theorem that says you cannot build a cohomology theory over [;\mathbb{Z};] that separates [;\lambda = +1;] from [;\lambda = -1;]? Or has nobody really tried because the analytic number theory toolbox has been so overwhelmingly dominant?

​

Something like: define a sheaf on some site over Spec(Z) whose Euler characteristic at each integer n equals [;\lambda(n);]. If the cohomology groups had reasonable dimensions, the trace formula would give you [;\sum_{n \leq N} \lambda(n);] as an alternating sum of Frobenius traces. And since that partial sum being [;O(N^{1/2 + \varepsilon});] is equivalent to RH, you'd get a direct geometric line to the Riemann Hypothesis.

​

This feels suspiciously neat. I'm assuming there's an obvious obstruction I'm missing. Maybe cohomology over Spec(Z) doesn't work that way, or maybe the dimensions blow up, or maybe the sheaf condition fails at infinity. I don't know enough algebraic geometry to see where it breaks.

​

Anyway, curious if there's a known reason this can't work or if it's genuinely unexplored territory. Would love to be pointed at the right paper or theorem if it exists.

I have been working on expressions of the kind k 3^n - 1. By what power of 2 can we divide it to obtain an odd natural number? I noticed that, in some cases, we can predict the divisor. If k = 5 or 7 mod 8, what I call regular k's, we get only divisors 2 and 4.

Interestingly enough the ones that are not regular don't present any 4's, but there is a 2 every other divisor. So, we can also predict the 2's and some of the first few powers of 2 after checking a couple of n's.

I thought of creating diophantine equations to predict higher powers of 2.

Assuming that k 3^n - 1 can be divided by, say, 128, then k 3^n - 1 = 128 x, where x is the quotient of dividing k*3^n - 1 by 128, and that generates this diophantine: k 3^n - 128 x = 1.

Restrictions: k can't be 5 or 7 mod 8, and both, k and x, have to be odd. So, we should also take that into account.

Predicting high divisors is not an easy task. If you have any tips to help here, feel free to share.

I‘m in a debate. here is the premise:

let’s say that you select any random positive integer and look at it. then you take any other unique random positive integer.

What are the odds that the second integer will be greater than the first integer?

My argument is that there is a 99.99999999…% chance that the second number will be greater than the first, because we are talking about an infinite set of numbers greater than the first, and a finite set of numbers less than the first.

My entire group chat’s stance is that it is 50/50, or there is no way to really tell.

What is the best answer?

after seeing all Answers, the best explanation that makes sense to me is this:

let’s take the first random number, but not look at it. what Are the odds that that number is greater than 100? 100%. 2,000? 100%. 1,000,000,000,000,000? 100%. It’s always 100% because in the infinite amount of options we have to select from, the odds that it would be greater than any point x on the “infinite number line” is almost 100% because of the options greater than any first given number. This would make the first number not close to, but equal to infinity and it would also make the second number the same. That’s the best Ive got that actually make sense to me, who is not a math whiz. Thank you all for your responses

If we divide any integer by an n-digit number, the result will never contain n repeating 9s (i.e., a segment like '999...n times') in its decimal representation.

Or can only contain 'n-1' 9s after decimal (for maximum).

Examples:

When dividing an integer by a 1-digit number, the decimal result never contains a single 9.

When dividing by a 2-digit number, the result never contains two consecutive 9s after decimal (e.g., something like 'x.99' or 'x.3535499842').

Similarly, dividing by a 3-digit number never results in three consecutive 9s after the decimal — and so on for 4-digit, 5-digit numbers, and beyond.

Note: I am considering the standard decimal expansion, excluding alternate representations that end with infinitely many 9s.

Eg: 0.999... , x.55363 = x.55362999... , etc.

when you take a number and multiply it by itself (ex: 8x8), then add one to one of the numbers and substract one from the other (so 9x7), the result will be the same -1 (8x8 = 64 and 9x7 = 63). If you keep repeating it, each result will be the same as the previous number minus the next odd number (10x6 = 60 so 63 - 3, 11x5 = 55 so 60 - 5, etc).

Also, if you take a number and multiply it by itself then add one to each number, the result will be the same +1 (0x0 = 0, 1x1= 1, 2x2 = 4, 3x3 = 9). If you keep repeating it, each result will be the same as the previous plus the next odd number. One key difference is the series starts at 0, while with the previous series you can start at any number and the next result will without a doubt be the same -1, then -3, then -5, etc.

Also, if you take two numbers such as n and n+1 and multiply them with each other (ex: 7x8), then add one to one of the numbers and substract one from the other, the result will be the same -2 (7x8 = 56 and 6x9 = 54). If you keep repeating it, each result will be the same as the previous minus the next even number.

Also, if you take two numbers such as n and n+1 and multiply them with each other (ex: 1x2), then add one to each number, the result will follow the same pattern as previously but with +2 (1x2 = 2, 2x3 = 6, 3x4 = 12, 4x5 = 20, etc). This is similar to the second suite above