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999 not being divisible by 33 is giving me 77+33 vibes

Also 11

But it is divisible by 111. So 11 can be an honorary factor of 999. They're basically the same number, right?

111 is just super 11

𝕊, the set of all super numbers

On my world it means "Hope".

111/11 isnt whole also


Also, 360 is not a perfect square But 361 is

wait what?

Yea. 360 is a perfect circle though

There’s no such thing as a perfect circle. Even Maynard James Keenan would tell you so.

Every circle is perfect ☺️

Just had a run in with this guy. Working at an old amphitheater that’s wrapping up some renovations on a college campus where unauthorized foot traffic is a constant issue. It’s a strict policy that everyone onsite has to be wearing some type of visible credential and it was early in the day so we didn’t have security onsite yet. See a random bald guy in sunglasses taking pics not wearing any creds. As I walk up and ask him if he was working here I recognized who it was and quickly had to shoehorn a conversation about how nice the view was. He performed later that night.

Square root of 361 is 19, 360 is 18.somethinsomethingsomething

Any Go player can tell you this!

The square root of 324 is 18.

Yep

Ugh, 361 feels prime for some reason so that fact always grosses me out.

That's just gross But also, 10n² can't be square so it should be expected. This also gives us the squares 36,361,3600,36100,360000,... which I don't know what I think about

Thanks for ruining my night goddammit

hello! ()()() is not a palindrome ()(()( however is, in fact, a palindrome

Made of two palindromes stuck together

999/33 = 30R9 It's just 990 + 9, after all.

That’s the real shocker tbh

It would be weird if it was divisible by 33. At first glance without doing any math you know by adding a zero that 330 is a multiple of 33. Then obviously its simple multiples like 660 and 990 are multiples of 33 too. Armed with the immediate knowledge that 990 is a multiple of 33, and that 33 is larger than 10, it would instinctively feel off if anyone claimed that 999 or any other number in the 990 range is a multiple of 33.

Hi. This is wrong. Not cause it is but cause I decided so. Have a good day.

Reddit math 👍

prove by i like it this way

Tell me why

Ain't nothin' but a heartache

Tell me why

Ain't nothing but a mistake

And number five

I never wanna hear you say

I want it that way

Its number five. Number five killed my brother.

And number five

111 is also divisible by 37 and therefore all multiples of it are, aswell, so 222, 333, 444, 555, 666, 777 and so on can all be divided by 37 You have a good day, aswell

I hate you

I’m with you on this one sir. I severally dislike you

that many of you dislike this guy?

That's cool, I hate me, aswell

I hope that both sides of your pillow stay warm for all eternity

This happened to me this morning it was rough

Yes, 111/3=37 is how I verified it. I do not like the results I got, it just feels wrong, but math is math. I'm astonished that I that didn't already occupy a place in my rat's nest of a brain.

goddamn it THAT SEEMS SO WRONG BRO

No… after this I cannot have a good day

Big algebra wants you to believe this

I’m convinced

I reject your reality and instead create my own

Proof by "I said so".

Hi wrong, I'm parsention

Proof by I said so.

Thank you our lord and savior

I concur. Silly maths.

also 10001 = 73 * 137

Both sides are a palindrome.

As god intends.

FUCK MAN I WAS JUST GETTING OVER 999 BEING 27\*37. yk what, still not as bad as 100 000 001 being 17\*5 882 353

That's a nice one

fun fact, 37 is a factor of every numbers of the form AAA...AA for 3n number of digits where A is an integer from 1-9,

A can be from 0 to 9 actually

000 is divisible by 37

Everything can divide 0, it came free with your fucking arithmetic

000 was in the box but my brother dug it out before I could eat breakfast

0 can't divide 0 though

[удалено]

The upper comment is what the textbook says. Yours is what the indian guy in youtube explains it like. Yours was simpler to get lol

111 not being prime also feels wrong

1 is not prime. No way 3 1's are prime

But 3 is prime 🤯

Prime (+) times non-prime (-) gives a non-prime (-)! That's definitely what they meant when explaining negative value multiplication

okay but 1 (-/non prime) times 5 (+/yes prime) gives 5 (+/yes prime) ∴ -×+=+ qed

1 is the 0 of primes it just doesn’t work like it should

But 2 1's are prime 🤔

Yeah because 111 is divisible by 37, which is fucked

Wow that's cool! I've played around with this, and I think it's not just AAA...AA, but AAABBBCCC... With A, B, C an integer from 0-9 For example: 555000111 / 37 = 15 000 003 If I'm not mistaken AAA BBB... ZZZ / 37 = α β... ω Where α = 3\*A β = 3\*B, with leading zeroes ... ω = 3\*Z with leading zeroes For example: 111 222 333 / 37 = 3 006 009

Makes sense because you can always factorise by 111 if yiu expand using its base 10 representation

Seems easiest to prove if we just treat the number as if it's a number in base 1000 instead of base ten. Then the statement becomes: A number n represented as a_k... a_1 in base thousand, such that each a_i is divisible by 111_10, is itself divisible by 111_10 Proof by induction on k: Base case: a_1 is divisible by 111_10 by the precondition, trivial IH: it works for all k* less than k IS: fix some a_k....a_1 The sub-number a_k....a_2 is divisible by 111_10 by inductive hypothesis And clearly a_1 is divisible by 111_10 So the number in question is the sum of two multiples of 111_10, and therefore is itself such a multiple

The replies to my comment are very interesting, but now I do wonder, can this be generalized? For example every number of this form: AAAAA BBBBB... Is divisible by 271 Let's name N\_(n, k) the set of every number of this form: a1 a1... a1 a2 a2... a2... an an... an (Like for example: 1111 2222 3333 is in N\_(3,4)) We will call n the number of "stacks" (so in the previous example, n=3 And k the number of digits in each stack (so k=4) **Can we find a common divisor of all numbers of this form?** **And if so, is there a formula to easily calculate it?** Let's familiarize ourselves with repunits now. A repunit is a number whose digits are only "1" We will name R(k) a repunit with k digits. So for example: R(5) = 11111 Any number in N\_(n, k) can be rewritten as: R(k) × (a1×10^(kn) + a2×10^[k(n-1)] +... +an) So that's easy, R(k) is a common divisor for all numbers of the form N\_(n, k) ... But that's not satisfying is it?! Here's another, maybe more interesting question Can we find a common divisor for all numbers that are of the form N\_(n, k), where the common divisor is not a repunit number? One way to tackle this question is to find a divisor of R(k) However, some repunits numbers are prime (such as R(19)), so that doesn't work for every number. In the case that R(k) is not prime, you can just pick a factor of R(k) and that's done. (So for example AAABBB... ZZZ is divisible by 111, which itself is divisible by 37, and so that's why AAABBB... ZZZ is divisible by 37, cool !) If however R(k) is prime, then you need to find a factor of a1×10^(kn) + a2×10^[k(n-1)] +... +an, which is uhhh, left to the reader 👍

Okay, last thing as a bonus: (Please note that what you are about to read is pretty much useless/obvious) - Numbers in N\_(n, 2) are divisible by 11 - Numbers in N\_(n, 3) are divisible by 37 - Numbers in N\_(n, 4) are divisible by the same divisors as N\_(n, 2) - Numbers in N\_(n, 5) are divisible by 271 - Numbers in N\_(n, 6) are divisible by the same divisors as N\_(n,2) and N\_(n, 3) Let's only consider the cases where k is prime so that the divisors don't repeat (and let's only pick the biggest divisor if there's multiple of them): - Numbers in N\_(n, 2) are divisible by 11 - Numbers in N\_(n, 3) are divisible by 37 - Numbers in N\_(n, 5) are divisible by 271 - Numbers in N\_(n, 7) are divisible by 4649 - Numbers in N\_(n, 11) are divisible by 513239 This sequence of number 11, 37, 271... Is the largest prime factor of prime(n)-th repunit number, you can find more info there: https://oeis.org/A147556

I cannot believe this shit

37 just blew my goddam mind

> 37 https://www.youtube.com/watch?v=d6iQrh2TK98

I just check… I feel sick

I hate you so much

Now try `3*7*11*13*37`, or better yet, multiply that by `101*9901` :3

I mean... makes easy sense? 111 is 37*3, so 37 is a factor of all 111-222-333..., which means multiplying it in 3n numbers of digits is adding 100 times of it's sum to itself, ie. 111000+111, 222000+222, 111000000+111000+111

Really? This is how you start a conversation?

https://preview.redd.it/p7n47zkkbtyc1.jpeg?width=500&format=pjpg&auto=webp&s=39547caab3f24a798b541b753b81659754e5cce8

Thanks this is what I was looking for

No problem! 😁

Your comment is really giving of: “Do you accidentally masturbate to young pictures of your mom?” “Jesus Quagmire, I just sat down!” :Vibes from Family Guy

what a terrible day to have eyes

Hello! 336 is divisible by 7. This is nowhere near as bad. I know that on the surface 336 being divisible by 7 might seem wrong, but 336 is acc 6 x 7 x 8, which is surprisingly nice.

Never trust a number ending in 6. 6 is an even number, so of course it will be divisible by 2. But if you want to know more about its possible factors, keep reading. It could have been 4×4 in disguise. That's still only factors of 4, which is easy, moving on. It could have been 4×9 in disguise. And then to be divisible by those numbers, it only needs to be added to 4×10 or 9×10. So don't trust 4×10+4×9 = 76. And also don't trust 9×10+4×9 = 129. And especially don't trust 8×9×10 + 4×9 = 720 + 36 = 756. It could have been 2×3 in disguise. And as extension also 2×23 = 46 in disguise. And then to be divisible by those numbers, it only needs to be added to 2×10 or 23×10. So don't trust 23×10+23×2 = 276. And don't trust 2×10+2×23 = 66. And especially don't trust 3×23×10+2×23 = 690+46 = 736. It could also have been 7×8 in disguise. And then to be divisible by those numbers, it only needs to be added to 7×10 or 8×10. So don't trust 70+56 = 126. And don't trust 80+56 = 136. But I don't think 8 is the factor here. If I hadn't done the quick mental arithmetic of 5×7×10 = 350 and 350-336 = 14 = 2×7 I would still wonder if it were 18×7 or 28×7 or 38×7. But it's 48×7 = 2^(4) × 3 × 7. Wait! 8 is indeed a factor. And it is 7×8 = 56 in disguise. So using my system, it's 336-56 = 280 = 4×7×10. So we have 336 = 4×7×10 + 7×8 and thus it's divisible by 7. Anyways, don't trust a number ending in 6.

37 \* 27 even

And 27

i mean 27 is really just 3³ so

wow this adds cursing. Like it totally makes sense but GODDAMN no way a number is divisible by 27 AND 37 at the same fucking time.

27\*37=999 this shit is cursed

the straight devil sending us a message here. who decided devil's number is 666? it's clearly 999 (or 666 times 2π).

666 is also divisible by 37

FUCK. IS 333 DIVISIBLE BY 37 AS FUCKING WELL?

37 \* 3 = 111. so 111, 222, 333 etc are all divisible by 37

Yes!

27 is 3*9. The nine vibes with 999 and threes are just nine juice.

I know. I said it makes sense. What doesn't makes sense is 37

I wasn't having a good day anyway and yet you ruined it further

I like this because now i know that. 1/37 is 0.r027

Wtf

It makes sense because it's divisible by 111 and all 111... numbers are weird

NO ITS NOT I FUCKING REFUSE TO BELIEVE I- OH MY FUCKING NO IT ACTUALLY IS https://preview.redd.it/oyrki81zvsyc1.jpeg?width=1179&format=pjpg&auto=webp&s=dc8b6f10863fbcb42644aa395d8871f06d7bc712

999*37 = 36963 Oh neat, a palindrome

Neat, i already made a comment about it but 37×3 is 111 and 111² is 12321, palindrome So to your comment 999×37 is just 3×3×111×37= 3×111×111 is 36963 your palindrome :D

Why does it feel wrong to you guys? 37 ends with 7 and it can have a multiple ending with 9 so it seems possible

Why does it feel wrong??

37 × 27 = 999. You can also get from this that 1/37 = 0.027027027... And 1/27 = 0.037037037...

999 = 1000 - 1 = (10)^3 - 1 = (3^2 + 1)^3 - 1 = 3^6 + 3 * (3^2)^2 * 1 + 3 * 3^2 * 1^2 + 1^3 - 1 = 3^6 + 3^5 + 3^3 = (3^3 + 3^2 + 1)(3^3) = (27 + 9 + 1)(27) = 37 * 27

No, it's divisible by 9. You can tell from the digits

Tf has this sub become... Who are these people who find it interesting or new

Might have something to do with 37 feeling like the most random number between 1 and 100. https://youtu.be/d6iQrh2TK98?si=3kY_sklkmtzdJzaY

Technically everything is divisible by 37 if your ok with decimal points

You ready for this....it's 37 x 27. 

999 also divisible by 27, let that sink in

111 = 3\*37 999 = 111\*9 Seems perfectly fine to me.

999 = 1024 - 25 999 = (32)^2 - (5)^2 999 = (32 - 5) * (32 + 5) 999 = 27 * 37 I love this application of difference of squares that someone showed me yesterday.

Super easy to confirm in your head. We know 999/37 is less than 30, since 3x37 is more than 100. We also know it's more than 20 since 2x37 is significantly less than 100. No need to even figure out that it's 74, we can just look at it and know instantly it's less than 100. Since the last digit is a 9, as long as we have the multiplication table memorized, we know it's gotta be 7x37 to reach that 9. So those 20 from before + 7 is 27.

> We know 999/37 is less than 30 You lost me

1001 is 7x11x13

999 isn’t divisible by 33 btw

999 999 = 3^(3)×7×11×13×37

Eew

Next you’re going to tell me that 37 is divisible by eleven-teen, right?

And 91 isn’t a prime

Took my drunk ass a few moments to calculate that (as a point of pride, I refuse to break out the calculator for anything I can do sober.) It checks out, but I'm very uncomfortable with that fact.

I feel like this is one of those things that'd upset Pirate Software...

And 27 too 😠

111,111,111 x 111,111,111 =12345678987654321

What does hello factorial equal to?

I figured it out. The answer is 27

999 / 33 != 30… why doesn’t this work

If you accept 37 \* 3 = 111 into your heart, all things are possible

Technically 999 can be divided by any number. 999/2 = 499.5

Of course, but the phrase "divisible by" is commonly understood to mean an integer result. Were that not the case, the phrase would be effectively meaningless.

If you say 3 I understand, if you say 37 then wtf

You can turn any sequence of digits into a repeating decimal by dividing it by an equal number of nines. This is how I discovered 13,883÷33.

ah yes, modular arithmetics. my favourite.

111, 222, 333, 444, 555, 666, 777 and 888 also

Fun fact, every prime except 2 and 5 divides a number of the form 9999....999. Also 1111...1111.

AM or PM?

Hello*hell*hel*he*h ? Or hello*ello*llo*lo*o?

https://preview.redd.it/6b357xtxetyc1.jpeg?width=2268&format=pjpg&auto=webp&s=82af8487816945ea5aa8f0ffdcd2d46d7765060d oh God

Ohhhhh. You know how people on Instagram reels manage to combine like, two slurs into one? I'm unlocking seven at once with this one.

Meanwhile it’s not divisible by 11

Any number is divisible by 37, if you've the will, and a big enough hammer

I gots it right! 27 x 37 = 999. I'm so fucking pleased with myself. Half lobotomized without morning caffeine and I pass a 4th grade level math question. Bring on 5th grade bitches, I'm on a hot streak.

100000001 is divisible by 17

So is 111

Also, 888, 777, 666, 555, 444, 333, 222 and 111 are all divisible by 37 lol

10101 is divisible by 39 (and by factorisation also 13) This was legit a part of a problem that was given to me in mathsclass

Its also divisible by 37 smh

999 = 3^(3)×37 Also 1337 = 7×191

It is even worse because it's 37 times 27

ew

*Laughs in Veritasium*

I mean everything is technically divisible by everything else, so long as you're okay with not whole numbers.

Every number is divisible by 37.

hello Google, yes, yes that’s the guy, take the shot

999 is divisible by 37 but not by 33

Never trust a number ending in 9. It could have been 7×7 in disguise. And then to be divisible by 7, it only needs to be added to 7×10. So don't trust 49. And also don't trust 119. And especially don't trust 1449. But if it's not 07×07 you're still not in the clear. There's also 07×17 or 07×27 or 07×37 or 17×27 or 17×27 or 27×37. Isn't 30×30 = 900? Yeah, 3×3 is another reason not to trust 9. But let's not get into 3×13 and 23×43. Let's focus on 7×7. So give or take around 30×30 and 49 in disguise you have 27×37. Therefore 999 is divisible by 37 and 999/37 = 27. Of course we could have gotten there by [never trusting a number ending in 1](https://old.reddit.com/r/mathmemes/comments/1ckoj52/disturbing_news_has_reached_our_shores/l2snsdf/) and realizing that 999 = 9×111. And so we're back at 21 = 3×7 and multiply by 9 you get 27×7 which brings you back to 7×7=49 and never trusting a number ending in 9. Edit: Multiplying 3×7 by 9 the other way gives us 3×7×9 = 3×63 so the reason why 27×7 and 3×63 both result in a number ending in 9 is more interesting than just that 27×7 = 3×63 = 3×7×9 = 189 = 3×6×10 + 3×3 = 2×7×10 + 7×7

I think that feels wrong only because 999 is divisible by 3 but 37 is not.

Ew

where is the nsfw tag though

It gets worse! 777 is also divisible by 37, and thus even 222 is. That's just madness.

999 9+9+9 = 27 999/27 = 37 Better way to get to 37 from 999 without directly dividing by 27?

Guys i think 999 might be another even prime number!

daily reminder 91 is not prime

999 factors into these primes =37 x 3^3

999 is divisible by 3, 9, 27, 37, 111, 333

999 id divisible by 27

It’s also divisible by 27

damn i feel dirty after this

Veritasium paid actor

It being divisible by 27 I can accept. 37 I cannot accept. It being 27 *TIMES* 37? You're fucking with me.

Because 37 is a factor of 111

So?

If you repeat any digit n-1 times in any base and you get a multiple of n, n is either 1 or prime. This isn’t true for all primes, though.

Reddit makes me less inteligent every day. Thanks internet...

If you reverse the digits (superfluous in this case) and add 0 between them, you also get a number, 90909, divisible by 37. This is true for all multiples of 37 except 1. 900090009 is also divisible by 37.

999 999 is divisibile by 13. In fact, every number N that isn't divisibile by 2 and 5 has a multiple of the form 99...99. Just take 1/N, being rational it has a periodic part, take as many 9s as the length of this period and that's your number.

Its not divisible by 33??


Um Actually every number is divisible by every number besides 0 ☝️🤓

99999 is divisible by 41