1waffle1
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| Joined: 16 Oct 2007 |
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| 24 Dec 2011 01:08 AM |
I could never seem to find any of these. Are there any proofs about the relationship between the digits in numbers?
Ex. In base 10, the sum of the digits of any positive integer multiple of 9, mod 9, is always 0. |
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Quenty
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| Joined: 03 Sep 2009 |
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| 24 Dec 2011 01:11 AM |
| Yeah. But also in geometry. Try googling them. Or ask your math teacher. |
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GigsD4X
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| 24 Dec 2011 01:14 AM |
ab*11=a{a+b}b
Where a and b are digits of a 2-digit number and the content inside the braces represents the digit to be inserted into the number the braces are in. |
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1waffle1
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| 24 Dec 2011 02:13 AM |
What about this:
function collatz(x) return x==1 and x or collatz(x%2 == 1 and x*3 + 1 or x/2) end
It will never crash as long as x is a positive integer. o_o |
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| 24 Dec 2011 02:20 AM |
| 10 will always be the value of whatever base you're using. :D |
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1waffle1
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Oysi
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NXTBoy
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| Joined: 25 Aug 2008 |
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| 24 Dec 2011 09:26 AM |
@Oysi: No. You've mistated the problem:
x = k*9 (x_1 + x_2 + x_3 + x_4 + ... ) % 9 == 0
Where x_n denotes the nth digit in base 10 |
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sdfgw
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| 24 Dec 2011 09:37 AM |
@Oysi:
it's the _sum of the digits of_ the multiple of 9, not just the multiple of 9, that is 0 modulo 9
so you just have to prove that the sum of the digits of a multiple of 9 is also a multiple of 9?
uh
100a + 10b + c = 9x
in mod 9
99a + a + 9b + b + c = 0
= a + b + c = 0
I guess you could extend that to any number of digits, but I have no idea how you'd write that as a proof |
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Oysi
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sdfgw
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| 24 Dec 2011 10:12 AM |
@Oysi:
we learned it add maths camp (no seriously)
Modular arithmatic is funkeh because you can just add and take away multiples of the number you're modding (terminology?) by. e.g.
2 = 7 (mod 5) is a true fact.
If you take a four digit number in the form
1000a + 100b + 10c + d
then obviously its digital form is 'abcd' (e.g. 2000 + 500 + 30 + 4 = 2534)
if we're given that it's a multiple of 3, then
1000a + 100b + 10c + d = 3x
for some integer x
now, we know that 9, 99, 999 et cetera are all multiples of 3, so 999a, 99b, and 9c are also multiples of 3
now we enter the wonderful world of modulo 3
1000a + 100b + 10c + d = 3x (mod 3)
and we can take off those multiples, as we know that taking away a multiple of 3 from a number does not change whether or not it is a multiple of 3 (e.g. 8 is not a multiple of 3, so neither is 8 - 3 = 5)
a + b + c + d = 3x (mod 3)
so the sum of the digits of any 4 digit multiple of 3 is also a multiple of 3.
yeahso it's dark hax, and with a little intuition and fancy notation I guesses you could apply it to any number of digits, but that's above my education system
this stuff is non-curricular you see |
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Oysi
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stravant
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| Joined: 22 Oct 2007 |
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| 24 Dec 2011 10:48 AM |
| The reason it's harder to find proofs of this nature is because they're often quite difficult problems to construct real formal proofs for. It's easy to see that a lot of them work by inspection, but a lot harder to formally prove them. |
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sdfgw
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| 24 Dec 2011 10:50 AM |
100a + 10b + c = 5x
all multiples of 10 are also multiples of 5, so this reduces to
c = 5x
It's a rather delightful concept. You should try writing out multiplication tables with modular arithmetic. The patterns are splendid.
I recall there being some pattern with multiples of 11... I think it's
10000a + 1000b + 100c + 10d + e = 0 (mod 11)
a - b + c - d + e = 0
so if you alternately add and subtract the digits of a multiple of 11, you also get a multiple of 11. Uhh
70000 + 2000 + 400 + (0) + 2 = 72402
7 - 2 + 4 + 2 = 11
yaay
[In case you didn't pick up on it, you can also do things like
10 = -1 (mod 11)] |
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sdfgw
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| 24 Dec 2011 10:52 AM |
| @stravant: Isn't there a way of writing what I did, except for n number of digits (using some sum notation)? |
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stravant
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| 24 Dec 2011 11:11 AM |
With things you can just apply easily to every power of 10 like 5/10 multiple, it's not very hard. However, some other patterns can be surprisingly difficult to prove.
For instance, not totally related, but even proving that every infinite series of digits actually represents a real number is quite difficult, and requires a number of other proofs to even approach the question. |
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1waffle1
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| Joined: 16 Oct 2007 |
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| 24 Dec 2011 12:26 PM |
| What about the Collatz conjecture? It works. I went up to 500 million and it worked for every number. But I don't understand why. |
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sdfgw
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| 24 Dec 2011 12:33 PM |
| @1waffle1: There's no guarantee that it does work. It might fail for numbers not tested yet. |
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| 24 Dec 2011 12:39 PM |
| Perhaps I will prove it through indirect supposition :D I doubt that there's a way to prove this directly. |
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stravant
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| 24 Dec 2011 01:28 PM |
| There's always a way to prove something directly if it can be done indirectly. In fact, there's a while stream of math dedicated to proving stuff only directly, that is, without using proof by contradiction. |
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| 24 Dec 2011 01:57 PM |
@waffle:
Collatz _conjecture_. It remains an unproven conjecture in mathematics. Bonus points for you if you can prove it, but you'd have to learn yourself some number theory. |
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1waffle1
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| 24 Dec 2011 02:01 PM |
@stravant: That may or may not be true, I'll ask my geometry teacher, but it's not very practical in some cases. |
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nightname
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| 24 Dec 2011 02:06 PM |
@Stravant
That reminds of something. They used to think that all mathematical equations, and other mathematical dilemmas can all be simplified down to axioms. Then they came across math.sqrt(-1), and they solved it by giving it the number "i".
Now they have even newer problems, and people no longer think that everything can be simplified/solved.
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1waffle1
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| Joined: 16 Oct 2007 |
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