Quote:
Originally Posted by Marie Monday
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
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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