$\square+\square+\square=30$, with boxes filled using $1, 3, 5, 7, 9, 11, 13, 15$, possibly repeated. How? From the days I started to learn Maths, I've have been taught that 

Adding Odd times Odd numbers the Answer always would be Odd; e.g., 
  $$3 + 5 + 1 = 9$$

OK, but look at this question 

This question was solved and the answer was 30, how it was possible? Need a valid explanation please.
 A: You have answered your own question: the sum of an odd number of odd numbers must be odd. Therefore it cannot equal 30.
You should not believe everything you read in a photo on the internet.
A: Clearly impossible in base 10. Can it be done in another number base?
In base 5:  $11+11+3=30$
Actually there are many other possibilities!
Another possibility: fill the boxes with $\binom{5}{3}$,15 and 5.
It holds that $\binom{5}{3}+15+5=30$ (base 10). Notice that $\binom{5}{3}=10$. I can't see any rule being violated as I'm using the 2 parenthesis in the list of valid symbols provided in addition to the numbers 5, 3, 15 and 5 and no extra symbol.
A: Do you have to fill in all three boxes? I would just put 15 in two of the boxes, e.g., $$\fbox{ } + \fbox{15} + \fbox{15} = 30,$$ and hope people understand the first box as being an implicit 0.
A: 
you can also repeat the numbers

Wonder if that means $\,11,5+13,5+5=30\,$ (where the $\,,\,$ comma works as 
decimal separator).
A: If this is a riddle, I would do : 13,1 + 7,9 + 9
A: What about 
6+9+15=30?
they don't stay you have to fill the boxes with the numbers in their usual orientation, after all. :-)
A: *

*Change the base:  $$11_9 + 11_9 + 11_9 = 10+10+10 = 30  \text{.}$$

*Parentheses are in the list of usable box contents.  Commas too.  But you're upper limited to one pair of parens and to seven commas.
$$(15+15,15)+15 = 30  \\ (15+15,15+15) = 30  \text{.}$$  Here, the parens represent the GCD.
Edit:
Change of base can also be made to work if number repetition were eliminated.
$$  11_{5} + 15_{7} + 13_{9} = 6 + 12 + 12 = 30  \text{.}  $$
A: It's
$$3_3+3_3+3_3=30.$$
Where $3_3$ is read as 3 base 3.
