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Old 03-15-2021, 02:47 PM   #1 (permalink)
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Default 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:
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Old 03-15-2021, 03:06 PM   #2 (permalink)
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the maths you wrote here aren't incorrect, but I still think the problem isn't solvable as it is stated here
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Old 03-15-2021, 03:13 PM   #3 (permalink)
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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
If I find a possible solution I'll post it here.
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Old 03-15-2021, 03:14 PM   #4 (permalink)
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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
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Old 03-15-2021, 03:19 PM   #5 (permalink)
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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
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Old 03-15-2021, 03:27 PM   #6 (permalink)
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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
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Last edited by Marie Monday; 03-15-2021 at 05:28 PM.
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Old 03-15-2021, 03:44 PM   #7 (permalink)
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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 delault 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
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Old 03-15-2021, 03:48 PM   #8 (permalink)
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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
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I want to open a school for MB's lost boys and teach them basic coping skills and build up their self esteem and strengthen their emotional intelligence and teach them about vegetables and institutionalized racism and sexism and then they'll all build a bronze statue of me in my honor and my bronzed titties will forever be groped by the grubby paws of you ****ing whiny pathetic white boys.
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Old 03-15-2021, 03:50 PM   #9 (permalink)
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ah ok good. 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. 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

Last edited by Eleanor Rigby 14; 03-15-2021 at 03:59 PM.
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Old 03-15-2021, 03:56 PM   #10 (permalink)
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math problems are exciting. Even though this one got me feeling properly embarassed about how long it took me to realise the obvious
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I want to open a school for MB's lost boys and teach them basic coping skills and build up their self esteem and strengthen their emotional intelligence and teach them about vegetables and institutionalized racism and sexism and then they'll all build a bronze statue of me in my honor and my bronzed titties will forever be groped by the grubby paws of you ****ing whiny pathetic white boys.
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