Completely pointless math facts

[enderlord99]

If you have a timer that, for some reason, can be set for FF:DD:hh:mm:ss with no error checking, you can set it for up to 1368715689 seconds by using only nines. This is somewhere between 43 and 44 years, but I can't convert it to YY:MM:DD:hh:mm:ss time because Wolframalpha insists on rounding prematurely.

[factotum]

Um, what is FF here?

[enderlord99]

Um, what is FF here?

Fortnight Fortnight

[gomipile]

If you have a timer that, for some reason, can be set for FF:DD:hh:mm:ss with no error checking, you can set it for up to 1368715689 seconds by using only nines. This is somewhere between 43 and 44 years, but I can't convert it to YY:MM:DD:hh:mm:ss time because Wolframalpha insists on rounding prematurely.

I don't know what standard you use for dividing out the months, but skipping months I get 43 years, 146 days, 14 hours, 48 minutes, and 9 seconds. That's if I use exactly 365 days per year with no leap days or leap seconds, though.

[JNAProductions]

If you have a timer that, for some reason, can be set for FF:DD:hh:mm:ss with no error checking, you can set it for up to 1368715689 seconds by using only nines. This is somewhere between 43 and 44 years, but I can't convert it to YY:MM:DD:hh:mm:ss time because Wolframalpha insists on rounding prematurely.

I appreciate this utterly useless fact. :P

Unfortunately, I don't have any to share myself.

[Douglas]

e𝜋-𝜋 is extremely close to 20. So close, in fact, that 6 of its first 7 digits past the decimal point are 9. And the one digit that isn't 9 is actually not at the end of those 7 - it's a 0 right in the middle instead.

19.99909997919...

[Fyraltari]

i to the power of i is a real number.

[truemane]

There's a great SMBC comic about someone asking his partner to speak 'mathy' to him in a sexy way, that culminates with him declaring she's a "mathy, mathy girl" that's always made me laugh, and would be the perfect compliment to this thread.

But it's JUST explicitly sexual enough that I don't want to link it.

But, suffice it say, you're all mathy, mathy playgrounders.

[Peelee]

There's a great SMBC comic about someone asking his partner to speak 'mathy' to him in a sexy way, that culminates with him declaring she's a "mathy, mathy girl" that's always made me laugh, and would be the perfect compliment to this thread.

But it's JUST explicitly sexual enough that I don't want to link it.

But, suffice it say, you're all mathy, mathy playgrounders.

Funny, my first thought was the SMBC comic about math transactions that ended with the harmonic sequence.

[Vizzerdrix]

If you have a timer that, for some reason, can be set for FF:DD:hh:mm:ss with no error checking, you can set it for up to 1368715689 seconds by using only nines. This is somewhere between 43 and 44 years, but I can't convert it to YY:MM:DD:hh:mm:ss time because Wolframalpha insists on rounding prematurely.

To be fair, no one likes a premature rounder. :smallwink:

[Grey_Wolf_c]

Every prime number squared is one more than a multiple of 24 (https://www.youtube.com/watch?v=ZMkIiFs35HQ).

Grey Wolf

[Douglas]

Every prime number squared is one more than a multiple of 24 (https://www.youtube.com/watch?v=ZMkIiFs35HQ).

Grey Wolf
Except 2 and 3.

[Peelee]

Oh, I'd forgotten about the 37 tricks for a long time.

A.) any single digit number multiplied by 3, then by 37, will give a product of the original number three times. Eg (5)(3)(37)=555, or (9)(3)(37)=999. The double digits also kinda work this way but you dont want to read me describe it.

2.) if you multiply by anything that gives a three-digit product, then move the first digit of the product to the back of the product, the resulting number is evenly divisible by 37. Eg (5)(37) = 185. 851/37 = 23, or (11)(37) = 407. 74/37 = 2. Or (15)(37) = 555. 555/37 = 15. If you want to be cheeky about it.

[factotum]

A.) any single digit number multiplied by 3, then by 37, will give a product of the original number three times. Eg (5)(3)(37)=555, or (9)(3)(37)=999.

Er, yes, because multiplying by 3 and then 37 is the same as multiplying by 111? It would be rather more obvious if you just said any single digit number multiplied by 111 gives three times the original number.

[Kato]

Oh, I'd forgotten about the 37 tricks for a long time.

A.) any single digit number multiplied by 3, then by 37, will give a product of the original number three times. Eg (5)(3)(37)=555, or (9)(3)(37)=999. The double digits also kinda work this way but you dont want to read me describe it.

2.) if you multiply by anything that gives a three-digit product, then move the first digit of the product to the back of the product, the resulting number is evenly divisible by 37. Eg (5)(37) = 185. 851/37 = 23, or (11)(37) = 407. 74/37 = 2. Or (15)(37) = 555. 555/37 = 15. If you want to be cheeky about it.

The first factotum already debunked (explained) but the second is a neater trick. I'm curious if there is a way to prove it.. And why it only works for three digits but I can't answer off the top of my head.

[Fyraltari]

I've got it.

So we have x = 37y where x is a three digit number, that means x = 100a + 10b + c where a, b and c are single digits (between 0 and 9 included).
(in Peelee's first example, y = 5, x = 185, a = 1, b = 8 and c = 5)
Moving the first digit to third position means creating a number z (851) where
z = 100b + 10c + a = 10(x - 100a) + a (eliminating the hundreds, moving the dozens and units up and adding the new units)
z = 10x - 1000a + a
z = 10x - 999a (is 999 a multiple of 37? Yes 27*37 = 999)
z = 10*37y - 27*37a
z = 37(10y-27a)
851 = 37(50-27) = 37*23

So for this to work y must be greater (or equal to) 3 times a. (edit3 : since 10y-27a => 0 <=>y => 27/10 *a)

EDIT:

Every prime number squared is one more than a multiple of 24 (https://www.youtube.com/watch?v=ZMkIiFs35HQ).

Grey Wolf

Oooh, that's a nice one.

EDIT the return:
37*2 = 74, not a three digit number, 37*3 = 111 a three digit number, so for x to have three digits y must be between 3 and 27.

EDIT the trilogy:
37y = 100a + 10b + c
y = (100/37)a + (10/37)b + (1/37)c
100/37 = 2,7027... > 2 so in order for x to have three digits, y must be greater than 3 times a (3 included).

There. Proved.

I now await the easier, more beautiful proof.

[Peelee]

Er, yes, because multiplying by 3 and then 37 is the same as multiplying by 111? It would be rather more obvious if you just said any single digit number multiplied by 111 gives three times the original number.
I am not a smart man. Also, I haven't looked at that since grade school, so blah.


I now await the easier, more beautiful proof.

https://www.smbc-comics.com/comics/20130120.gif

[Fyraltari]

https://www.smbc-comics.com/comics/20130120.gif

https://qph.fs.quoracdn.net/main-qimg-a3b19e99fad0ee2c1d663d4aa65d8dd1

[Kato]

Ah, nicely done! I didn't think of substituting the 10b+c which is really stupid in hindsight.

So the trick boils down to... 37 being a divisor of 999? Which means it should also work for only 3 (obviously), 9 (also) and 27 (less boring).
I think I'll remember that trick, thanks!

[Fyraltari]

Ah, nicely done! I didn't think of substituting the 10b+c which is really stupid in hindsight.
Thank you! substituting 10b+c isn't actually needed as the important part is realizing that moving the first digit around means y = 10(x-100a) + a. I didn't streamline my reasoning before posting (as is evident by my multiple edits :smallredface:) and I started with decomposing the numbers into multiples of ten which is always a good idea when there's digit manipulation (in base ten).


So the trick boils down to... 37 being a divisor of 999? Which means it should also work for only 3 (obviously), 9 (also) and 27 (less boring).
In order for it to work with 27, y needs to be greater than (or equal to) 3.7 times a (37/10 = 3.7) for every x in the hundreds.
27y = 100a + u <=> y = (100/27)a + u/27
100/27 = 3.70370... so it should work. Not sure about 9 and 3, though.


I think I'll remember that trick, thanks!
I probably will too.

[Rockphed]

i to the power of i is a real number.

... is it? The complex number is eiπ/2. So ii = ei^2π/2 = e-π/2.

Okay, so it is.

[halfeye]

Except 2 and 3.

Yes. These are primes. A pattern that doesn't include them is not a pattern for all primes.

[Grey_Wolf_c]

Yes. These are primes. A pattern that doesn't include them is not a pattern for all primes.

The actual mathematician in the video calls them subprimes. I'll take their word over yours, random internet person.

Grey Wolf

[halfeye]

The actual mathematician in the video calls them subprimes. I'll take their word over yours, random internet person.

Grey Wolf

Yeah, I'm a random internet person, so is the person in the video. The definition of prime numbers is what it is, look it up sometime.

[Peelee]

The actual mathematician in the video calls them subprimes. I'll take their word over yours, random internet person.

Grey Wolf

Counterpoint: it seems the mathematician in the video is defining primes based on the them having a value of one more than a multiple of 24 after being squared in addition to the other requirements. Which, after a few quick and dirty Google searches, does not seem to be a standardized definition at all.

That said, I'm not a mathematician, and dude could be using a known definition I didn't uncover.

[Fyraltari]

A prime number by definition is an integer that is multiple of only two* integers 1 and itself. By that definition 2 and 3 are primes.

1 has been excluded from the primes because it doesn't share the most useful properties of the other primes, most importantly that every integer can be written as a unique product of primes. So it was either getting 1 off the list or adding "except 1" to the primes' properties. I don't know of any reason not to consider 2 and 3 primes and I would like to hear that person's argument for it.

[eggynack]

The actual mathematician in the video calls them subprimes. I'll take their word over yours, random internet person.

Grey Wolf
I mean, it was a joke. Cause, like, subprime mortgages or whatever. The thing being proved was just that primes over three have this quality. Mathematicians do things like that sometimes, proving claims only with regard to almost all elements of a set, rather than all of them. The people in the video know that two and three are not prime.

[Peelee]

The people in the video know that two and three are not prime.

:smalleek:

[jayem]

I mean, it was a joke. Cause, like, subprime mortgages or whatever. The thing being proved was just that primes over three have this quality. Mathematicians do things like that sometimes, proving claims only with regard to almost all elements of a set, rather than all of them. The people in the video know that two and three are not prime.

Also the proof was actually (and explicitly) for all non-multiples of 2 and 3. But that would be boring and give away the trick.
And from X being co-prime with 2 and co-prime with 3 to generally prime seems a nice way of doing it. You've only lost composites of higher primes, so 25, 35, 49... so for a region it nearly works both ways by 1000 there are 166 primes out of 333 and by 5000 it's 667 out of 1666. But now you need weasel words for excluding 2&3) IIRC correctly 2 kind of gets this a fair bit, but it's so easy to say odd-prime, and of course odd, non three prime.gives away the pleading.


Anyhow for my fact
60*60*24*365 is within 1% of pi*107 or sqrt(10)*107. Allowing a moderate compromise between order of magnitude and accurate numbers for anything involving years.

[factotum]

... is it? The complex number is eiπ/2. So ii = ei^2π/2 = e-π/2.

Okay, so it is.

But all real numbers are just a sub-set of complex numbers anyway...

[eggynack]

:smalleek:
At some point you're going to have to accept that this arbitrary pun designation for 2 and 3 is so widely beloved in the mathematics community that these numbers have been removed from the list of primes to make the joke work better. Math is 95% objective rigor, and 5% jokes about real estate.

[Jay R]

The original statement is not a property of primes in the first place. The square of any number not divisible by 2 or 3 is 1 more than a multiple of 24

252 = 625 = 624 + 1 = 24*26 + 1
352 = 1225 = 1224 + 1 = 51*24 + 1
492 = 2401 = 2400 + 1 = 100*24 + 1

Any number not divisible by 2 or 3 is either 6n - 1 or 6n + 1

(6n - 1)2 = 36n2 -12n + 1 = 12(3n2 - n) + 1

(6n + 1)2 = 36n2 + 12n + 1 = 12(3n2 + n) + 1

If n is even, then so is 3n2, so 3n2 - n and 3n2 + n are both even.
If n is odd, then so is 3n2, so 3n2 - n and 3n2 + n are both even.

Therefore in either case, it's 12 * an even number + 1, so it's one more than a multiple of 24.

[Peelee]

The square of any number not divisible by 2 or 3 is 1 more than a multiple of 24

Any number not divisible by 2 or 3, you say?:smallamused:

[jayem]

Any number not divisible by 2 or 3, you say?:smallamused:
Positive integer, I think works, then (and I think we've avoided a phantom 'not' ?)
I was going to exclude 12. But that is 0*24+1, so I don't need to

2,3,4 are of course divisible by 2 or 3
52=1*24+1
6=3*2,
72=2*24+1

8=2*4, 9=3*3, 10=2*5
112=5*24+1
12=3*4
132=7*24+1

14=2*7,15=3*5,16=4*4
172=12*24+1
18=3*9
192=15*24+1

232=22*24+1
252=5*5*5*5=625=26*24+1

I'd assumed (having not looked into it) it would have gone in adjacent values of 24, which it clearly doesn't.

[Peelee]

Positive integer, I think works, then (and I think we've avoided a phantom 'not' ?)
I was going to exclude 12. But that is 0*24+1, so I don't need to

Nah, I was thinking of 1 but then forgot to think of multiplying by zero. Hoisted on my own petard!

[Craft (Cheese)]

The original statement is not a property of primes in the first place. The square of any number not divisible by 2 or 3 is 1 more than a multiple of 24

252 = 625 = 624 + 1 = 24*26 + 1
352 = 1225 = 1224 + 1 = 51*24 + 1
492 = 2401 = 2400 + 1 = 100*24 + 1

Any number not divisible by 2 or 3 is either 6n - 1 or 6n + 1

(6n - 1)2 = 36n2 -12n + 1 = 12(3n2 - n) + 1

(6n + 1)2 = 36n2 + 12n + 1 = 12(3n2 + n) + 1

If n is even, then so is 3n2, so 3n2 - n and 3n2 + n are both even.
If n is odd, then so is 3n2, so 3n2 - n and 3n2 + n are both even.

Therefore in either case, it's 12 * an even number + 1, so it's one more than a multiple of 24.

Since all primes greater than or equal to 5 are not divisible by 2 or 3 either, that means it is indeed a property of these numbers.

And if that seems overly pedantic then maybe mathematics isn't for you :smalltongue:

[Peelee]

Oh, I just remembered another math fact: 1 is the largest number.

http://www.smbc-comics.com/comics/1485876000-20170131.png

SMBC is an academic journal, right?

[Unavenger]

https://imgs.xkcd.com/comics/approximations.png

[Jay R]

Any number not divisible by 2 or 3, you say?:smallamused:

Well, the mere fact that we are considering divisibility at all limits it to the integers. But yes, any integer not divisible by 2 or 3.

(-1)2 = 1 = 0(24) + 1
(-17)2 = 289 = 12(24) + 1


Since all primes greater than or equal to 5 are not divisible by 2 or 3 either, that means it is indeed a property of these numbers.

Yes, the rest of the primes is a subset of the set of numbers that have that property. But given that we've proven that many non-primes have that property, and two primes don't have that property, and showed that the property is not dependent on the primeness of the primes who have it, it's reasonable to say that it's not a property of the primes in the first place.

For example, 11, 131, 373, 757, and 10,301 are prime numbers that are palindromes in base 10. And there are many others (although 11 is the only one with an even number of digits). That does not make being a palindrome in base 10 a property of the primes.


And if that seems overly pedantic then maybe mathematics isn't for you :smalltongue:

Well, don't tell the university that awarded my Ph.D., or my statistics and algebra students, or the companies I do statistical consulting work for.

[eggynack]

Yes, the rest of the primes is a subset of the set of numbers that have that property. But given that we've proven that many non-primes have that property, and two primes don't have that property, and showed that the property is not dependent on the primeness of the primes who have it, it's reasonable to say that it's not a property of the primes in the first place.

For example, 11, 131, 373, 757, and 10,301 are prime numbers that are palindromes in base 10. And there are many others (although 11 is the only one with an even number of digits). That does not make being a palindrome in base 10 a property of the primes.
I disagree. I think that being able to make a claim for almost all elements of a set, meaning that only finitely many elements do not have that quality, is sufficient for talking about the set in connection with the quality. This is, I must assume, not the case for this palindromic quality (and I wouldn't be overly surprised were there finitely many such cases), so being a palindrome in base 10 is not a quality of the primes where this essentially is.

[ben-zayb]

The decimal forms for 1/7, 2/7, 3/7, 4/7, 5/7, and 6/7, are just cyclic permutations of each other that contain the digits 142857.

Conversely, the first 6 multiples of 142857 are just cyclical permutations of itself.

[georgie_leech]

I disagree. I think that being able to make a claim for almost all elements of a set, meaning that only finitely many elements do not have that quality, is sufficient for talking about the set in connection with the quality. This is, I must assume, not the case for this palindromic quality (and I wouldn't be overly surprised were there finitely many such cases), so being a palindrome in base 10 is not a quality of the primes where this essentially is.

It's not a useful definition though, because it doesn't describe primes, but "numbers not divisible by 2 or 3." Yes, any prime number except for 2 or 3 will meet that definition, but... well, all squares are rectangles, not all rectangles are squarts. We can't use the definition of a superset to define a subset in particular. Math Fact: every Prime Number turns out to be a Number.

[Caerulea]

Positive integer, I think works, then (and I think we've avoided a phantom 'not' ?)
I was going to exclude 12. But that is 0*24+1, so I don't need to

2,3,4 are of course divisible by 2 or 3
52=1*24+1
6=3*2,
72=2*24+1

8=2*4, 9=3*3, 10=2*5
112=5*24+1
12=3*4
132=7*24+1

14=2*7,15=3*5,16=4*4
172=12*24+1
18=3*9
192=15*24+1

232=22*24+1
252=5*5*5*5=625=26*24+1

I'd assumed (having not looked into it) it would have gone in adjacent values of 24, which it clearly doesn't.
I see you that list, and raise you the first 1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 numbers.
1^2 = 1 = 24 * 0.0 + 12 = 2 * 1.0
3 = 3 * 1.0
4 = 2 * 2.0
5^2 = 25 = 24 * 1.0 + 1
6 = 2 * 3.0
7^2 = 49 = 24 * 2.0 + 1
8 = 2 * 4.0
9 = 3 * 3.0
10 = 2 * 5.0
11^2 = 121 = 24 * 5.0 + 1
12 = 2 * 6.0
13^2 = 169 = 24 * 7.0 + 1
14 = 2 * 7.0
15 = 3 * 5.0
16 = 2 * 8.0
17^2 = 289 = 24 * 12.0 + 1
18 = 2 * 9.0
19^2 = 361 = 24 * 15.0 + 1
20 = 2 * 10.0
21 = 3 * 7.0
22 = 2 * 11.0
23^2 = 529 = 24 * 22.0 + 1
24 = 2 * 12.0
25^2 = 625 = 24 * 26.0 + 1
26 = 2 * 13.0
27 = 3 * 9.0
28 = 2 * 14.0
29^2 = 841 = 24 * 35.0 + 1
30 = 2 * 15.0
31^2 = 961 = 24 * 40.0 + 1
32 = 2 * 16.0
33 = 3 * 11.0
34 = 2 * 17.0
35^2 = 1225 = 24 * 51.0 + 1
36 = 2 * 18.0
37^2 = 1369 = 24 * 57.0 + 1
38 = 2 * 19.0
39 = 3 * 13.0
40 = 2 * 20.0
41^2 = 1681 = 24 * 70.0 + 1
42 = 2 * 21.0
43^2 = 1849 = 24 * 77.0 + 1
44 = 2 * 22.0
45 = 3 * 15.0
46 = 2 * 23.0
47^2 = 2209 = 24 * 92.0 + 1
48 = 2 * 24.0
49^2 = 2401 = 24 * 100.0 + 1
50 = 2 * 25.0
51 = 3 * 17.0
52 = 2 * 26.0
53^2 = 2809 = 24 * 117.0 + 1
54 = 2 * 27.0
55^2 = 3025 = 24 * 126.0 + 1
56 = 2 * 28.0
57 = 3 * 19.0
58 = 2 * 29.0
59^2 = 3481 = 24 * 145.0 + 1
60 = 2 * 30.0
61^2 = 3721 = 24 * 155.0 + 1
62 = 2 * 31.0
63 = 3 * 21.0
64 = 2 * 32.0
65^2 = 4225 = 24 * 176.0 + 1
66 = 2 * 33.0
67^2 = 4489 = 24 * 187.0 + 1
68 = 2 * 34.0
69 = 3 * 23.0
70 = 2 * 35.0
71^2 = 5041 = 24 * 210.0 + 1
72 = 2 * 36.0
73^2 = 5329 = 24 * 222.0 + 1
74 = 2 * 37.0
75 = 3 * 25.0
76 = 2 * 38.0
77^2 = 5929 = 24 * 247.0 + 1
78 = 2 * 39.0
79^2 = 6241 = 24 * 260.0 + 1
80 = 2 * 40.0
81 = 3 * 27.0
82 = 2 * 41.0
83^2 = 6889 = 24 * 287.0 + 1
84 = 2 * 42.0
85^2 = 7225 = 24 * 301.0 + 1
86 = 2 * 43.0
87 = 3 * 29.0
88 = 2 * 44.0
89^2 = 7921 = 24 * 330.0 + 1
90 = 2 * 45.0
91^2 = 8281 = 24 * 345.0 + 1
92 = 2 * 46.0
93 = 3 * 31.0
94 = 2 * 47.0
95^2 = 9025 = 24 * 376.0 + 1
96 = 2 * 48.0
97^2 = 9409 = 24 * 392.0 + 1
98 = 2 * 49.0
99 = 3 * 33.0
100 = 2 * 50.0
101^2 = 10201 = 24 * 425.0 + 1
102 = 2 * 51.0
103^2 = 10609 = 24 * 442.0 + 1
104 = 2 * 52.0
105 = 3 * 35.0
106 = 2 * 53.0
107^2 = 11449 = 24 * 477.0 + 1
108 = 2 * 54.0
109^2 = 11881 = 24 * 495.0 + 1
110 = 2 * 55.0
111 = 3 * 37.0
112 = 2 * 56.0
113^2 = 12769 = 24 * 532.0 + 1
114 = 2 * 57.0
115^2 = 13225 = 24 * 551.0 + 1
116 = 2 * 58.0
117 = 3 * 39.0
118 = 2 * 59.0
119^2 = 14161 = 24 * 590.0 + 1
120 = 2 * 60.0
121^2 = 14641 = 24 * 610.0 + 1
122 = 2 * 61.0
123 = 3 * 41.0
124 = 2 * 62.0
125^2 = 15625 = 24 * 651.0 + 1
126 = 2 * 63.0
127^2 = 16129 = 24 * 672.0 + 1
128 = 2 * 64.0
129 = 3 * 43.0
130 = 2 * 65.0
131^2 = 17161 = 24 * 715.0 + 1
132 = 2 * 66.0
133^2 = 17689 = 24 * 737.0 + 1
134 = 2 * 67.0
135 = 3 * 45.0
136 = 2 * 68.0
137^2 = 18769 = 24 * 782.0 + 1
138 = 2 * 69.0
139^2 = 19321 = 24 * 805.0 + 1
140 = 2 * 70.0
141 = 3 * 47.0
142 = 2 * 71.0
143^2 = 20449 = 24 * 852.0 + 1
144 = 2 * 72.0
145^2 = 21025 = 24 * 876.0 + 1
146 = 2 * 73.0
147 = 3 * 49.0
148 = 2 * 74.0
149^2 = 22201 = 24 * 925.0 + 1
150 = 2 * 75.0
151^2 = 22801 = 24 * 950.0 + 1
152 = 2 * 76.0
153 = 3 * 51.0
154 = 2 * 77.0
155^2 = 24025 = 24 * 1001.0 + 1
156 = 2 * 78.0
157^2 = 24649 = 24 * 1027.0 + 1
158 = 2 * 79.0
159 = 3 * 53.0
160 = 2 * 80.0
161^2 = 25921 = 24 * 1080.0 + 1
162 = 2 * 81.0
163^2 = 26569 = 24 * 1107.0 + 1
164 = 2 * 82.0
165 = 3 * 55.0
166 = 2 * 83.0
167^2 = 27889 = 24 * 1162.0 + 1
168 = 2 * 84.0
169^2 = 28561 = 24 * 1190.0 + 1
170 = 2 * 85.0
171 = 3 * 57.0
172 = 2 * 86.0
173^2 = 29929 = 24 * 1247.0 + 1
174 = 2 * 87.0
175^2 = 30625 = 24 * 1276.0 + 1
176 = 2 * 88.0
177 = 3 * 59.0
178 = 2 * 89.0
179^2 = 32041 = 24 * 1335.0 + 1
180 = 2 * 90.0
181^2 = 32761 = 24 * 1365.0 + 1
182 = 2 * 91.0
183 = 3 * 61.0
184 = 2 * 92.0
185^2 = 34225 = 24 * 1426.0 + 1
186 = 2 * 93.0
187^2 = 34969 = 24 * 1457.0 + 1
188 = 2 * 94.0
189 = 3 * 63.0
190 = 2 * 95.0
191^2 = 36481 = 24 * 1520.0 + 1
192 = 2 * 96.0
193^2 = 37249 = 24 * 1552.0 + 1
194 = 2 * 97.0
195 = 3 * 65.0
196 = 2 * 98.0
197^2 = 38809 = 24 * 1617.0 + 1
198 = 2 * 99.0
199^2 = 39601 = 24 * 1650.0 + 1
200 = 2 * 100.0
201 = 3 * 67.0
202 = 2 * 101.0
203^2 = 41209 = 24 * 1717.0 + 1
204 = 2 * 102.0
205^2 = 42025 = 24 * 1751.0 + 1
206 = 2 * 103.0
207 = 3 * 69.0
208 = 2 * 104.0
209^2 = 43681 = 24 * 1820.0 + 1
210 = 2 * 105.0
211^2 = 44521 = 24 * 1855.0 + 1
212 = 2 * 106.0
213 = 3 * 71.0
214 = 2 * 107.0
215^2 = 46225 = 24 * 1926.0 + 1
216 = 2 * 108.0
217^2 = 47089 = 24 * 1962.0 + 1
218 = 2 * 109.0
219 = 3 * 73.0
220 = 2 * 110.0
221^2 = 48841 = 24 * 2035.0 + 1
222 = 2 * 111.0
223^2 = 49729 = 24 * 2072.0 + 1
224 = 2 * 112.0
225 = 3 * 75.0
226 = 2 * 113.0
227^2 = 51529 = 24 * 2147.0 + 1
228 = 2 * 114.0
229^2 = 52441 = 24 * 2185.0 + 1
230 = 2 * 115.0
231 = 3 * 77.0
232 = 2 * 116.0
233^2 = 54289 = 24 * 2262.0 + 1
234 = 2 * 117.0
235^2 = 55225 = 24 * 2301.0 + 1
236 = 2 * 118.0
237 = 3 * 79.0
238 = 2 * 119.0
239^2 = 57121 = 24 * 2380.0 + 1
240 = 2 * 120.0
241^2 = 58081 = 24 * 2420.0 + 1
242 = 2 * 121.0
243 = 3 * 81.0
244 = 2 * 122.0
245^2 = 60025 = 24 * 2501.0 + 1
246 = 2 * 123.0
247^2 = 61009 = 24 * 2542.0 + 1
248 = 2 * 124.0
249 = 3 * 83.0
250 = 2 * 125.0
251^2 = 63001 = 24 * 2625.0 + 1
252 = 2 * 126.0
253^2 = 64009 = 24 * 2667.0 + 1
254 = 2 * 127.0
255 = 3 * 85.0


The decimal forms for 1/7, 2/7, 3/7, 4/7, 5/7, and 6/7, are just cyclic permutations of each other that contain the digits 142857.

Conversely, the first 6 multiples of 142857 are just cyclical permutations of itself.
I never noticed that. Cool.

—Caerulea

[Peelee]

Math Fact: every Prime Number turns out to be a Number.

I'm gonna need to see a proof of that.

[tiornys]

Assume p is prime. Then p belongs to P, the set of all primes. By definition, P is a subset of N, the set of all natural numbers. Therefore p contained in P implies p contained in N. By definition, p an element of N implies p is a number. QED

Yeah, yeah, what it really boils down to is "p is a number by definition" but this bloated phrasing amused me.

[eggynack]

It's not a useful definition though, because it doesn't describe primes, but "numbers not divisible by 2 or 3." Yes, any prime number except for 2 or 3 will meet that definition, but... well, all squares are rectangles, not all rectangles are squarts. We can't use the definition of a superset to define a subset in particular. Math Fact: every Prime Number turns out to be a Number.
Why can't we use a quality of a superset to define a subset? That seems pretty arbitrary. Your sarcastic math fact is a math fact nonetheless, and if, at some point, a proof relating to the primes relies on the numerical quality of prime numbers, then you'll be happy to know that prime numbers are indeed numbers. None of the math facts in this thread are especially useful, for such was the demand made of the facts, but sometimes you just want to talk about primes, and in such cases applying the quality to the subset could make sense.

That primes have this quality is an obvious consequence of the fact that numbers divisible by neither two nor three have this quality, but basically any mathematical claim is necessarily a direct deductive extrapolation from some claimed axioms such that just stating the axioms would be logically equivalent. The only standard you're applying, then, is how "obvious" the fact is, which is a rather subjective thing.

[halfeye]

Why can't we use a quality of a superset to define a subset?

Because logic doesn't work like that.

"All dogs are mammals" does not imply that "all mammals are dogs".


Assume p is prime. Then p belongs to P, the set of all primes. By definition, P is a subset of N, the set of all natural numbers. Therefore p contained in P implies p contained in N. By definition, p an element of N implies p is a number. QED

Yeah, yeah, what it really boils down to is "p is a number by definition" but this bloated phrasing amused me.

Except, that might be wrong. Infinity might be a prime, and infinity is not a number.

[eggynack]

Because logic doesn't work like that.

"All dogs are mammals" does not imply that "all mammals are dogs".

Logic works exactly like that. You just kinda did it backwards. Mammal is a superset of dog, so any quality of mammals as a set is also a quality of dogs as a set. You can't use a quality of a subset to define a superset, however. Thus, the quality of dogs that they are dogs does not apply outwards to mammals, but the quality of mammals that they are mammals does apply inwards to dogs. For a less confusing example, the "warm blooded" quality of mammals also applies universally to all dogs, but the "never lives to 40" quality of dogs doesn't apply outwards to all mammals.



Except, that might be wrong. Infinity might be a prime, and infinity is not a number.
How could infinity possibly be a prime?

[georgie_leech]

Logic works exactly like that. You just kinda did it backwards. Mammal is a superset of dog, so any quality of mammals as a set is also a quality of dogs as a set. You can't use a quality of a subset to define a superset, however. Thus, the quality of dogs that they are dogs does not apply outwards to mammals, but the quality of mammals that they are mammals does apply inwards to dogs. For a less confusing example, the "warm blooded" quality of mammals also applies universally to all dogs, but the "never lives to 40" quality of dogs doesn't apply outwards to all mammals.




Right, which means we can't use this property to judge whether a number is prime or not. Squaring the number and seeing if it is one more than a multiple of 24 only tells you whether it's divisible by 2 or three. If the square is 1 more than a multiple of 24, then neither is a divisor of the base; if it isn't, than one or both of 2 and 3 are divisors. It's a fact that becomes increasingly minor in comparison to the divisors it can't have. Ruling out 2 potential prime factors for, say, 23495823 isn't a very useful way of determining anything.

It would be more like defining a dog as "not a reptile." Great, that's already implied by the mammal bit, it's a lousy test.

[halfeye]

How could infinity possibly be a prime?

How could anyone prove that it isn't (I'm not saying someone hasn't, there have been a lot of very clever people playing with infinity, just that I don't remember hearing of a proof)?

[georgie_leech]

How could anyone prove that it isn't (I'm not saying someone hasn't, there have been a lot of very clever people playing with infinity, just that I don't remember hearing of a proof)?

I've never heard someone claim a plus sign or pi was a prime number either. Just because it's a mathematical object doesn't mean it belongs to the class of objects that can be described as a prime number. Hint: infinity isn't a number.

[eggynack]

Right, which means we can't use this property to judge whether a number is prime or not. Squaring the number and seeing if it is one more than a multiple of 24 only tells you whether it's divisible by 2 or three. If the square is 1 more than a multiple of 24, then neither is a divisor of the base; if it isn't, than one or both of 2 and 3 are divisors. It's a fact that becomes increasingly minor in comparison to the divisors it can't have. Ruling out 2 potential prime factors for, say, 23495823 isn't a very useful way of determining anything.

It would be more like defining a dog as "not a reptile." Great, that's already implied by the mammal bit, it's a lousy test.
I mean, yeah, it doesn't operate in that particular context, a method of showing a number prime or not. That's not necessarily the only context it could work in though. Maybe you just want to prove a new thing about the prime numbers, and a necessary piece of information in the proof is that their squares are like this or whatever.


How could anyone prove that it isn't (I'm not saying someone hasn't, there have been a lot of very clever people playing with infinity, just that I don't remember hearing of a proof)?
I just have no idea how the notion of prime would even connect to the notion of infinity. Like, a prime number is divisible by one and itself, but no other numbers. So, infinity would have to be divisible by one, but not divisible by two. I have to think it's impossible to come up with a self-consistent definition of infinity that has both those qualities.

[halfeye]

I've never heard someone claim a plus sign or pi was a prime number either. Just because it's a mathematical object doesn't mean it belongs to the class of objects that can be described as a prime number. Hint: infinity isn't a number.

Someone up-thread said that all primes were numbers. Then Peelee suggested that might not be true. I'm just running with that idea to see how far it goes. Obviously infinity isn't a number, since that's the whole point of the exercise, do primes have to be numbers? For that matter, is zero prime? is zero a number?

I suggest that the obvious doesn't apply to infinity. Infinity seems a lot more like a number than a plus sign does, and it's much more likely to be an integer than pi, which is clearly not an integer, maybe there's a proof that infinity is not an integer, that sounds like the sort of thing I'd never have heard of, or remembered if I did.


I just have no idea how the notion of prime would even connect to the notion of infinity. Like, a prime number is divisible by one and itself, but no other numbers. So, infinity would have to be divisible by one, but not divisible by two. I have to think it's impossible to come up with a self-consistent definition of infinity that has both those qualities.

Why is it impossible to do that? If you can prove that divisibility by two is a necessary feature of infinity then we are done, but I'm not seeing it yet.

[Peelee]

For that matter, is zero prime?

No; right off the bat, it's not divisible by itself.

ETA: I never cast aspersions on all primes not being numbers, I just embraced the fact that proofs can and will be performed for the most obvious of statements. Of course, I could always revert to the ol' "there are no prime numbers because 1 is the largest number," if I wanted to get stupid with it. :smallwink:

[factotum]

Infinity divided by any non-infinite number is infinity, by definition. You could argue that means infinity is divisible by all numbers, or by none, depending on your point of view.

[Jay R]

How could anyone prove that it isn't (I'm not saying someone hasn't, there have been a lot of very clever people playing with infinity, just that I don't remember hearing of a proof)?

Infinity isn't a specific number. It's just shorthand for a limit.

"1/∞ = 0" really means "As x grows without bound, the value of 1/x approaches 0".

Primeness is a property of specific numbers.

[Fyraltari]

Right, which means we can't use this property to judge whether a number is prime or not. Squaring the number and seeing if it is one more than a multiple of 24 only tells you whether it's divisible by 2 or three. If the square is 1 more than a multiple of 24, then neither is a divisor of the base; if it isn't, than one or both of 2 and 3 are divisors. It's a fact that becomes increasingly minor in comparison to the divisors it can't have. Ruling out 2 potential prime factors for, say, 23495823 isn't a very useful way of determining anything.

It would be more like defining a dog as "not a reptile." Great, that's already implied by the mammal bit, it's a lousy test.

The OP isn’t asking for ‘completely pointless’’ facts. A simple test to identify primes would be extremely pointful.

[Jay R]

A completely pointless math fact.

1,729 is the smallest natural number that is the sum of two cubes, two different ways.

13 + 123 = 1 + 1,728 = 1,729
103 + 93 = 1,000 + 729 = 1,729

[Grey_Wolf_c]

A completely pointless math fact.

1,729 is the smallest natural number that is the sum of two cubes, two different ways.

13 + 123 = 1 + 1,728 = 1,729
103 + 93 = 1,000 + 729 = 1,729

It was also the cab number of a visitor to Ramanujan's sick bed.

Grey Wolf

[Craft (Cheese)]

The harmonic series (1 + 1/2 + 1/3 + 1/4 + ...) diverges to infinity, unless you throw out all denominators containing a 9, in which case it converges to ~22.92.

(There's nothing particularly special about 9: It works with any digit, though the converging value you get will be different.)

[jayem]

The harmonic series (1 + 1/2 + 1/3 + 1/4 + ...) diverges to infinity, unless you throw out all denominators containing a 9, in which case it converges to ~22.92.

(There's nothing particularly special about 9: It works with any digit, though the converging value you get will be different.)

That is quite cool, it seems odd that it's such an extreme difference, Is there a neat 'trick' to calculating the convergence.

[crayzz]

That is quite cool, it seems odd that it's such an extreme difference, Is there a neat 'trick' to calculating the convergence.

I don't remember the trick, but it's something to with with the fact that most numbers have any particular digit in them.

[danzibr]

(-1/2)!=√π

There’s a cool one with limits of the zeta function around 1 basically canceling to give you the value gamma, hard to write on a phone though.

[Craft (Cheese)]

I don't remember the trick, but it's something to with with the fact that most numbers have any particular digit in them.

If you pick a random 100-digit number, the likelihood that it doesn't contain a 9 is (8 * 9 ^ 99) / (9 * 10 ^ 99), or about 0.000026. That probability approaches zero as the number of digits increase, so "almost all" numbers have a 9 in them somewhere.

What's perhaps a bit more surprising is that this also works for any arbitrary string of digits, and thus any harmonic series with any string of digits removed from the denominator will also converge to a finite value. Such as 42, 1337, 8675309, or the first 10 trillion digits of pi.

[TrashTrash]

The letter x is used more times in mathematics than it is used in the entire English language.

Due to the stupidity of subtracting negative numbers [i.e. 4-(-4)] you can just add like the negative's a positive [Example: 4-(-4) = 4+4]. Which means that working through the act of subtracting negatives is utterly pointless.

If all the digits in any number add up to 9, it's a multiple of 9. (108, 234, and 153 are odd but valid examples.)

[enderlord99]

If all the digits in any number add up to 9, it's a multiple of 9. (108, 234, and 153 are odd but valid examples.)

Actually, the first two are even but valid examples.

[tonberrian]

As a result of the above, all numbers with the digits of their decimal representation added up equal to a multiple of 3, are themselves multiples of 3.

[Kato]

Actually, the first two are even but valid examples.

I feel the desire to punch you for that one :smalltongue:

But those two are not pointless (well, not totally) and I thought pretty well known. Just more obscure cases similar to 'if a number ends on 0 (or 5) it's divisible by 10 (5)'

[Peelee]

The letter x is used more times in mathematics than it is used in the entire English language.


Well, it's used as a multiplication symbol, a variable placeholder, while in English it's used as a letter. So you are correct, two uses is indeed greater than one. :smalltongue:

[Fyraltari]

Well, it's used as a multiplication symbol, a variable placeholder, while in English it's used as a letter. So you are correct, two uses is indeed greater than one. :smalltongue:

Oh yeah, I remember when we used to have a symbol for multiplication. Aaaaah, youth.

[Khedrac]

Well, it's used as a multiplication symbol, a variable placeholder, while in English it's used as a letter. So you are correct, two uses is indeed greater than one. :smalltongue:


Oh yeah, I remember when we used to have a symbol for multiplication. Aaaaah, youth.

Actually I blame the problem on computers.

To me
x = the letter
and
× = the multiplication symbol

Similarly
/ is a slash
÷ is a division symbol
and
. is a full stop (or period)
· is a decimal point
What's worse is that I can't really do a proper subtraction sign on a computer, the closest is the n-dash – but they are supposed to be different, it definitely shouldn't be a hypen -. (I think the n-dash may actually be a subtraction operator and not an n-dash - it should be a + without the vertical and it does appear to be, suggesting that the n-dash is wrong...)

However the first typewriters used a limited characterset and computers followed. Ebven the QWERTY keyboard layout was designed to slow down typists who were jamming the mechanism of manual typewriters (which is why people are always coming up with faster keyboards where a moderately competent user is aster than an expert on QWERTY) but once QWERTY became standardised it became impossible to shift the mass market.

[Yet we still have problems when a computer decides it has a "US keyboard" plugged into it and one has to type in a password that is now different on the proper European keyboard actually attached.
Did you know that to Windows £¬| etc. are not counted as special characters when checking password security? They count for length but are not checked as aphanumeric nor as special characters...]

[tiornys]

If all the digits in any number add up to 9, it's a multiple of 9.


As a result of the above, all numbers with the digits of their decimal representation added up equal to a multiple of 3, are themselves multiples of 3.


But those two are not pointless (well, not totally) and I thought pretty well known. Just more obscure cases similar to 'if a number ends on 0 (or 5) it's divisible by 10 (5)'
Along these lines:

Any number whose last digit is even is a multiple of 2.
Any number whose last 2 digits are divisible by 4 is a multiple of 4. Furthermore, a number is divisible by 4 if the 2nd to last digit is even and the last digit is divisible by 4, OR if the 2nd to digit is odd and the last digit is even but NOT divisible by 4.
Any number whose last 3 digits are divisible by 8 is a multiple of 8. Furthermore, a number is divisible by 8 if the 3rd to last digit is even and the remaining 2 digits are divisible by 8, OR if the 3rd to last digit is odd and the remaining 2 digits are divisible by 4 but NOT divisible by 8.
This pattern continues to generalize across powers of 2. Any number whose last n digits are divisible by 2^n is a multiple of 2^n. If the nth to last digit is even, the number is divisible by 2^n if the remaining n-1 digits are divisible by 2^n, whereas if the nth to last digit is odd, the number is divisible by 2^n if the remaining n-1 digits are divisible by 2^(n-1) but not by 2^n.

[Excession]

What's worse is that I can't really do a proper subtraction sign on a computer, the closest is the n-dash – but they are supposed to be different, it definitely shouldn't be a hypen -. (I think the n-dash may actually be a subtraction operator and not an n-dash - it should be a + without the vertical and it does appear to be, suggesting that the n-dash is wrong...)

The symbol is − or Unicode 2212. It is separate from the n-dash. With WinCompose (https://github.com/samhocevar/wincompose) (and other compose key implementations) the sequence is - - =

Unicode contains almost every character you can think of, and even a few where nobody can work out where they came from :smallwink:

[factotum]

However the first typewriters used a limited characterset and computers followed. Ebven the QWERTY keyboard layout was designed to slow down typists who were jamming the mechanism of manual typewriters (which is why people are always coming up with faster keyboards where a moderately competent user is aster than an expert on QWERTY) but once QWERTY became standardised it became impossible to shift the mass market.

Entirely the opposite, actually--the QWERTY layout was designed to speed up typing, because what tended to jam the mechanism on early typewriters was when two adjacent letters were used in quick succession, and the QWERTY layout moves things around so that letters which commonly appear next to each other in English words aren't adjacent on the typewriter anymore. This is why the layout changes in different languages (e.g. AZERTY for French) because the letters which commonly appear next to each other in words changes.

[Yora]

The letters in a German keyboard are identical to QWERTY, except that the Z and Y are switched. Y is a virtually useless letter in the German language and appears almost exclusively in the place name "Bayern" and the family name "Meyer", the only letter more rare being X. But Z is the third rarest letter so there really was no point to this switch.
I also can not explain while all other special characters on a German keyboard are completely different from an English keyboard. What did that accomplish?!


e𝜋-𝜋 is extremely close to 20. So close, in fact, that 6 of its first 7 digits past the decimal point are 9. And the one digit that isn't 9 is actually not at the end of those 7 - it's a 0 right in the middle instead.

19.99909997919...
Doesn't the one 0 make the nines that come after it pretty much irrelevant as a measure of how close it is? It's practically 19.9991.


https://qph.fs.quoracdn.net/main-qimg-a3b19e99fad0ee2c1d663d4aa65d8dd1

This is my favorite math cartoon, I've not seen it in ages and it didn't seem like something you'd find in an internet search. I've just been thinking about it a few days ago.

[jayem]

Doesn't the one 0 make the nines that come after it pretty much irrelevant as a measure of how close it is? It's practically 19.9991.

Technically yes (by the more conventional metric), however it's a bit like the 'subprime' case.
By the Hanning Metric (how closely spelt things are) 19.99919999 is practically 19.9999999 (which of course is maximally different from 20.0000000)

For the sake of making a one in a thousand chance look like one in a million (and making it more memorable). It's totally worth the evil mixing of things rhetorically.



The 9's rule (at least in the more general case), I'd put as quite a bit more obscure than the 5&10's. There is a similar rule for 11's (subtract alternating digits).
On a slightly similar theme IIRC there are 4 numbers in every 100 where the last 2 digits of the square are the same as in the initial number X00, X01, X25, X76

[Douglas]

Doesn't the one 0 make the nines that come after it pretty much irrelevant as a measure of how close it is? It's practically 19.9991.
From a math perspective, yes. From a reading/writing perspective, no. I read a story once about someone doing a prank on some students who were doing an "implement a calculator" type of programming assignment by telling them that it's exactly 20 (or maybe it was "rounds to" 20), and that evaluating this was a standard test of floating point math algorithms. They took a while to figure it out, and I imagine they skimmed right over that 0 without noticing quite a few times.

[Fyraltari]

Actually I blame the problem on computers.

To me
x = the letter
and
× = the multiplication symbol

Similarly
/ is a slash
÷ is a division symbol
and
. is a full stop (or period)
· is a decimal point
What's worse is that I can't really do a proper subtraction sign on a computer, the closest is the n-dash – but they are supposed to be different, it definitely shouldn't be a hypen -. (I think the n-dash may actually be a subtraction operator and not an n-dash - it should be a + without the vertical and it does appear to be, suggesting that the n-dash is wrong...)

However the first typewriters used a limited characterset and computers followed. Ebven the QWERTY keyboard layout was designed to slow down typists who were jamming the mechanism of manual typewriters (which is why people are always coming up with faster keyboards where a moderately competent user is aster than an expert on QWERTY) but once QWERTY became standardised it became impossible to shift the mass market.

[Yet we still have problems when a computer decides it has a "US keyboard" plugged into it and one has to type in a password that is now different on the proper European keyboard actually attached.
Did you know that to Windows £¬| etc. are not counted as special characters when checking password security? They count for length but are not checked as aphanumeric nor as special characters...]
I meant that it has been a long time since I had to wrtite down a multiplication of two actual numbers rather than stuff like "PV = nRT" or "20(4 - y)² + 34x² = 27".

[crayzz]

From a math perspective, yes. From a reading/writing perspective, no. I read a story once about someone doing a prank on some students who were doing an "implement a calculator" type of programming assignment by telling them that it's exactly 20 (or maybe it was "rounds to" 20), and that evaluating this was a standard test of floating point math algorithms. They took a while to figure it out, and I imagine they skimmed right over that 0 without noticing quite a few times.

I think you're thinking of this XKCD comic (https://m.xkcd.com/217/).

[Peelee]

I think you're thinking of this XKCD comic (https://m.xkcd.com/217/).

I think you're thinking of this SMBC comic (https://www.smbc-comics.com/comic/2013-06-05)!

[Fat Rooster]

The sum digits to get mod 9 is definitely not pointless, and actually generalises to any number. For any number d you can build a sequence by taking 10i mod d. This sequence will eventually start to repeat, so this step is always finite. You can then calculate any number mod d by multiplying each digit in turn (starting from the right) by that term in the sequence, and summing them.

9 and 3 are special cases because their sequences are 1,1,1,1,1... The number 11 has a sequence 1,10,1,10... or equivilantly, 1,-1,1,-1... The numbers 5 and 2 have sequences 1,0,0... The number 7 is the first awkward one, going 1,3,2,6,4,5, and repeating those 6.

2,5,10: 1,[0]
3,9: [1]
4: 1,2,[0]
6: 1,[4]
7: [1,3,2,6,4,5]
8: 1,2,4,[0]
11: [1,10] (equivilantly [1,-1])
12: 1,10,[4]

[enderlord99]

Please stop adding "m." to the beginning of xkcd URLs.

[Peelee]

Please stop adding "m." to the beginning of xkcd URLs.

The phrase you're looking for is "please remove the "m.".

[crayzz]

Please stop adding "m." to the beginning of xkcd URLs.

Nah. I try to be mindful of that, but, if I'm posting links from my phone, mobile links are gonna sneak through eventually.

[Douglas]

I think you're thinking of this XKCD comic (https://m.xkcd.com/217/).
Yes, that's probably it, which means it's probably a made up joke. Bit of a shame, really.

[TrashTrash]

Well, it's used as a multiplication symbol, a variable placeholder, while in English it's used as a letter. So you are correct, two uses is indeed greater than one. :smalltongue:

That works :smallbiggrin: I was thinking more along the lines of the sheer number of equations that use 'x' as an unknown variable versus the amount of words with the letter 'x' in them in English. Yours is simpler though, I like it.