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Re: Math problems
Posted: June 17th, 2017, 10:14 am
by Zaknafein
Coco wrote:Now Im trying to figure out a way to sum all even (and only even) numbers via some formula, even if input is odd number. So if input is 7 for example the output should be 2+4+6=12. If input is 6,the result should be the same.
You can try to solve 2 + 4 + 6 + 8 + ... + 2n with the knowledge you already have about 1+2+...+n.
There is a hint:
1+2+ 4 + ...+n = S ---> 2 + 4 + 6 + ... + 2n = 2*1 + 2*2 + 2*3 + ... + 2*n = 2(1 + 2 + 3 + ... + n) = 2S
Coco wrote:Now Im trying to figure out a way to sum all even (and only even) numbers via some formula, even if input is odd number.
You can extend the previous solution to define your f(N), distinguishing if N is odd or even.
Coco wrote:Ratios between sums of even and odd numbers, their progression
You can define g(N) = 1 + 3 + ... + N, if N is odd. cause you have f(N) + g(N) = S, and you know what is g and S.
Same way defining g(N), if N is even.
Once you have g, you can define h(N) = f(N) / g(N), function of ratios.
You can also define a
k = h(k), for k=1,2,3,... b
k = f(k) - g(k), for k=1,2,3,...
Re: Math problems
Posted: June 17th, 2017, 12:13 pm
by David
Norbert wrote:
Code: Select all
fResult = ((float)iNumber / (float)iDivide);
if (IsInt (fResult))
Um, you could use the % (modulo) operator instead...
Anyway, writing a program for this seems to be overkill.
Re: Math problems
Posted: June 17th, 2017, 5:17 pm
by Coco
@Zaknafein
Are we still talking mathematical way? Program should be easy.
There is a hint:
I can be dumb sometimes...
Is it:
Also, if I multiply entire sequence by 2, I think the my nth N should then be reduced for half its size, and that can be a problem if its odd number.
As for ratios. I've no idea how to calculate them based on N. Although I know how they progress, I don't know how to put that on paper. Here is it:
Lets take a look at sequence part from 2 to 8:
Value of N even/odd ratio
2 2/1
3 1/2
4 3/2
5 2/3
6 4/3
7 3/4
8 5/4
What I've noticed is that ratio gets reciprocal value, then for next number, flip it again and add 1 both to numerator and denominator. One ore interesting thing is that with every next number (lets take evens for now, but it applies to both) nominator gets (-1) value of that number. Example:
4 ----- 3/2
Notice number 3 in nominator 4-3=1
6 ----- 4/3
again 6-4=2
8 ----- 5/4
and again 8-5=3
But I've no Idea how to explain this mathematically
PS
I just googled Menorca
Re: Math problems
Posted: June 20th, 2017, 12:25 am
by Coco
Norbert had spoken:
"even for very simple programming tasks I often look up things."
So is it advisable to learn programming that way? I'm having trouble just to write a function for sorting and inversing simple number lists. In python, where syntax couldn't be possibly easier
Sure there are sort() functions, and I've found the work behind it, but writing it myself...well... I have a vague idea how to do it, but I just can't complete it.
Re: Math problems
Posted: June 20th, 2017, 8:30 pm
by Zaknafein
Coco wrote:Are we still talking mathematical way? Program should be easy.
Yep, but sometimes you can't prove a property about an infinite set like integers.
Coco wrote:2n*(n+1)/2 => n(n+1) ?
You are right. You can prove also that 1+3+5+...+(2n-1)=n^2.
1+3+5+...+(2n-1) = (1+0)+(2+1)+(3+2)+...+( n + (n-1) ) = (1+2+3+...+n) + (0+1+2+...+(n-1) ) = [n(n+1)/2] + [(n-1)n/2] = (n^2 + n + n^2 - n)
/ 2 = (2n^2)/2 = n^2
Coco wrote:Notice number 3 in nominator 4-3=1
I think you mean that 2n-[Nominator
n] = n-1.
You can simplyfy (2+4+...(2n))/(1+3+...+(2n-1)), simplifying [n(n+1)] / [n^2] = (n+1)/n.
n+1 have no prime factor common with n (it doens't exist any prime p that divides (n+1) and n).
Finally, (n+1) is the simplest nominator, and the difference is (2n)-(n+1)=n-1.
Coco wrote:I just googled Menorca
You have to travel to Menorca at least 1 time
Re: Math problems
Posted: June 20th, 2017, 8:32 pm
by Zaknafein
By the way, is there any way to use Latex here?
Re: Math problems
Posted: June 21st, 2017, 9:07 am
by Coco
The nominators (and denominators) are behaving almost as an integer sequence. But they are getting a step behind with each step the number gets bigger. Between 2,4,6... the nominators of ratios are 2,3,4. A more interesting things are denominators in ratios. Check them out. Starting from number 2 they go 1,2,2 ,3,3,4,4
(just saw that pattern now)
I dont think latex can be used here, but Norbert certainly knows more about that.
Re: Math problems
Posted: June 21st, 2017, 10:28 am
by Norbert
Zaknafein wrote:By the way, is there any way to use Latex here?
Well, the official LaTeX forum (latex.org/forum) also uses the phpBB forum software, and they use
JavaScript code to present LaTeX content in new browser tabs using
Overleaf. Example link in a
post here. But unless you expect to use lots of LaTeX on this forum, it might be easier (for the forum admin; currently Falcury) if forum users use a
pastebin for mathematicians, such as
texpaste.com. Combined with pasting the code on this forum inside...
[/spoiler][/code]
...just in case the pastebin used shuts down.
Re: Math problems
Posted: June 24th, 2017, 3:40 pm
by Zaknafein
I tried to use pastebin to write the following problem, with
[/spoiler][/code] using also texpaste.com, but it doesn't work. I'll try to use Overleaf to write non elemental math problems. At the moment, there is a problem:
Let Susan and Paul two players.
Let 1<a<=b<100 integers.
Susan knows S=a+b.
Paul knows P=ab.
They also know the initial situation.
They have that conversation (they tell the truth and are intelligent enough):
P: I don't know a and b values.
S: I knew that you couldn't know that.
P: Now, I know a and b values.
S: I also know the values of a and b.
The problem is to find a and b. Good luck!
Re: Math problems
Posted: June 24th, 2017, 9:14 pm
by Norbert
Zaknafein wrote:I tried to use pastebin to write the following problem, with
[/spoiler][/code] using also texpaste.com, but it doesn't work.
You should be able to use something like
$$ E=mc^2 $$
in texpaste.com without problems.
Link:
http://www.texpaste.com/n/bvbkcxq2
Backup:
Re: Math problems
Posted: June 25th, 2017, 5:45 pm
by Coco
The only thing I can think of is that A and B can have range:
A = [2, 49]
B = [49, 97]
depending on the other number
EDIT:
My bad. B can also range from [2,97] if A=2
Re: Math problems
Posted: June 25th, 2017, 6:04 pm
by Coco
Actually, scratch all that. I wrote on paper that a+b<=100 xD
Re: Math problems
Posted: June 25th, 2017, 9:02 pm
by Zaknafein
Coco wrote:The only thing I can think of is that A and B can have range:
A = [2, 49]
B = [49, 97]
I think you are wrong, because A=3, B=53 is not a solution.
S=3+53=56.
P=3*53=159.
P is product of 2 prime numbers. Then, this is the only way to factorize 159. Paul should know the solution of (A,B), but he doesn't.
I think you need to make a program to solve it
Re: Math problems
Posted: June 25th, 2017, 9:11 pm
by Zaknafein
Here is another similar problem:
https://www.teamten.com/lawrence/puzzles/daughters.html
It can give you a hint about how to solve the previous problem.
Re: Math problems
Posted: June 25th, 2017, 10:38 pm
by Coco
36= 2*3*6 I haven't checked the solution.
So only thing I know is that:
A>1
A <= B
and
B<100
?
A and B are integers.
I don't really have any restrictions here. I've no idea how much is a+b or a*b, nor is there a limit of their sum/product. Doesn't this mean I can just pick any number for their values between 2 and 99 ofcourse?