The Official Math Thread
 

Any type of question, problem or coment related with the marvellous world of mathematics is welcome.
I'm going to start with a problem I have to solve but I can't.
It's the following one:
Juan says to Luisa: If you give me 3 coins I will have "n" times as many as you'll have. Luisa responds:
True, but if you give me "n" coins, then I will have three times what you have left. For what values of "n" are these statements true?
If you know how to solve it, please help.
I've already started solving it, but I can't finish it:

the maths you wrote here aren't incorrect, but I still think the problem isn't solvable as it is stated here :(

I think I'm about to finish it. Not sure if the solutions will be correct but I gotta try. It's not a compulsory thing, it's a competition, but it is sort of compulsory for me :laughing:
If I find a possible solution I'll post it here.

actually, they say values in plural, so could it be that they want you to express n in terms of x or y? that seems unlikely to me because I guess they'd tell you explicitly if they want that, but idk

Hell, if this ain't correct i'm gonna cry
P.S. some things are in SPanish
Edit: I got confused and wrote m instead of n somewhere, i gotta change that
But i think in general it's right

yoo I got the answer. omg jesus I swear to god normally I'm way better and quicker with these things. I just thought this was a default kind of math problem that it's not

edit: what you're doing here is the right method, I haven't checked whether all the maths details are correct
edit 2: so I had a look, and why are you splitting the fraction in two at some point? I think when you're saying n+13 >= 3n - 1 you're neglecting the +4 term you have

Idk if I understood the question correctly what I meant is that because x,y and n have to be natural numbers. Also, y=4+ n+13/3n-1. For "y" to be a natural number, n+13/3n-1 also has to be a natural number. I'm not neglecting the 4 cause while calculating "y" I give a natural value below 7 to "n" calculate the n+13/3n-1 and add 4.
P.S. What I need is an Official English Beginners Thread :laughing:

But actually in your case n+13/3n-1 can be a negative integer up to -4, so it doesn't have to be a natural number

What I had got after some algebra was: x = (7n+3)/(3-n)
which needs to be a natural number, like you said. Then, since the numerator is always positive the denominator needs to be positive too, so n < 3. Then both n=1 and n=2 work
we got different results though, but that may be me because I'm sloppy with algebra

I got 3 and 7 as possible results too. :eek: I'm going to check again if they fit as solutions.
P.S. Getting excited at a Math problem lol
Edit: They can be possible solutions.
I just replaced x+3=n (y-3) and 3(x-n)=y+n with the numbers obtained.
If n is 3 and both Juan and Luisa have 6 coins:
6+3=3(6-3)
and
3(6-3)=3+6
If n is 7 and Juan has 11 coins and Luisa 5:
11+3=7(5-3)
and
3(11-7)=5+12

math problems are exciting. Even though this one got me feeling properly embarassed about how long it took me to realise the obvious