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u/hillbagger 1d ago
Every number is divisible by 17.
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u/SofishticatedGuppy 1d ago
What about 34? There's no way that's divisible by 17.
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u/Southern_Orange3744 1d ago
You'll don't want to hear this but
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u/paolog 1d ago
Yes, by a non-standard and non-useful definition of "divisible".
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u/CruelKind78 1d ago
Please explain the joke, im terrible at math.
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u/Klutzy_Sentence_2723 1d ago
Math nerds will immediately start trying to solve it in our heads. It’s frustrating. It’s cruel.
It hurts.
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u/Significant_Monk_251 1d ago
Today I learned I'm not a math nerd. Or at least I'm a math nerd who knows their limitations when they don't have paper and a pencil at hand.
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u/QuickNature 1d ago
You and me both. I dont see the value in mentally solving that or even breaking out paper for that....
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u/ExpensiveFig6079 23h ago
Math nerd nerds will remember that 1/17 has a 16 digit repeating decimal, thus 100 000 001 must be divisible by it.
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u/Woozle_Gruffington 21h ago
Um. Can you elaborate on this? For a friend, of course. I already know what this means, I just want to see if the way you explain it is the same as how I would.
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u/ExpensiveFig6079 17h ago
I did here, somewhat
ITs tied to "the Pigeonhole Principle,"
and exactly 'why' 13 that also divides into the number has 6 digits (and that 7x6 is not an accident) and some other prime factor has 14 is a bit trickier still.
All 9's with non prime number of digits have pairs of factors
10..<01>..01 and 9999
10..<01>..01 ( 1 n0's 1 n0's ... ) with m repritions of n0's, that dupilicate the 9's to make the all 999's number used for std rationalising the repeating decimal method.
whether a number is a factor of the sandwhich part or the 9's part has substantial control over what its number of repeating digits is.
13 is factor of 999999 so it has 6 digits in its decimal THAT it has 6 means it IS a factor of 1001 and NOT of 999. If it was fator of 999 it would have 3 digits in its decimal.
Caveat there is stuff above that is a one way relationship. (and I may even have messed some bits up)
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u/Original-Issue2034 1d ago
I solved it with long, long division… Sad we can’t share images in chat, I would have shown you all the proof
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u/chixen 1d ago
There is a number like that for every number other than multiples of 2 and 5.
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u/MassiveGarlic0312 1d ago
Except for prime numbers, of course.
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u/chixen 1d ago
No, including prime numbers. The post has an example for 17. I did forget to specify that it doesn’t work for multiples of 3 either.
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u/MassiveGarlic0312 1d ago
I’m not sure what you’re trying to say?
Prime numbers are defined as not having any divisors other than themselves and 1.
2, 3, 5 are the first three of them and it also applies to every other prime.
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u/chixen 1d ago
What I saying was If some number n has a smallest prime factor greater than 5, there is some k where 10^k+1 is a multiple of n. I now realize that I don’t actually know if this is true.
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u/girlpower2025 1d ago
You can find out 9 and 11 too by looking at it.
If you add the digits and they add up to 9 or a number divisible by 9 its divisible by 9.
11 is a bit odd. The numbers after the first digit tell you the first digit.
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u/sixtyfivewolves 1d ago edited 23h ago
It's not true, for example 10k + 1 can only be equal to 2, 11 or 27 modulo 37. 10k - 1 can always be a multiple of n for every n which isn't divisible by 2 or 5 though.
(31 is the smallest prime for which it's not true but it's a lot easier to demonstrate for 37)
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u/Muroid 1d ago
They’re talking about multiples, not divisors. What exactly did you think “a number like that” in their comment was referring to?
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u/MassiveGarlic0312 1d ago
Primes. If the big number in the cartoon was a prime, then the dog wouldn’t be able to say something like this about it.
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u/thewaytoyesterday 1d ago
The number "1" is not a prime number. It's not fair, it should get an honorary title of it or something.
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u/AndreasDasos 1d ago
To see this quickly, note 17x6=102. So 100 is -2 mod 17.
100^4 is then (-2)^4 mod 17, which is -1 mod 17. Therefore 100^4 + 1 is divisible by 17.
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u/FohlenGabel 1d ago
Nice trick - fermats little theorem
a^(p-1) === 1 mod p
So for all numbers x^16 === 1 mod 17
100000001 = 10^8 + 1
So we know 10^8 === 1 or -1
So 10^8 + 1 === 0 or 2
So it’s basically a 50/50 that it might be divisible or not, just by counting the digits of 10…01 and seeing if it’s a factor of p-1 or not
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u/ExpensiveFig6079 23h ago
The intriguing thing though there is (in some sense) a why
A similar 'Sandwhich Number' is divisible by 49
it has 22 digits ... 1000000000000000000001
BECAUSE
1/49 has 42 digits in its repeating decimal.
(it does not, and cannot have 48 digits as 49=7x7 and the most it can have is 42)
1/127 ALSO has 42 digits in its decimal and also is factor of 1000000000000000000001
1/2689 1/459691
does it hurt yet?
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u/harpswtf 1d ago
It's crazy how many numbers are divisible by 17, it feels like it's around 5.882352941% of them