It would be interesting to know exactly what the examiners had in mind. A lot hinges on what is meant by "using the numbers".

if we can use mod, here is one of infinite possibilities:

1357911 mod 13 + 111315 mod 13 + 13571 mod 13 = 30

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If you add 3 odd numbers the answer cannot be even.  Therefore they cannot add up to 30

Q: The Question is simple: _ + _ + _ = 30

A: 1₍₅₎ + 11₍₅₎ + 13₍₅) = 30₍₅₎

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Nice try Badinage, but the question specifies odd numbers and 11₅ and 13₅ are even numbers!

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The comment "(Odd numbers upto 15)" looks like the poster's interpretation designed to mislead, rather than part of the problem specification.  That's what I'm claiming!! 

But the number 15 is specified and there is no 15₅.

I like your lateral thinking though!  Try it with numbers to the base 7 (make sure the they are all odd).

.

There is significant ambiguity here (admittedly, that's what often makes math puzzles interesting!).

"Numbers to be used: ..." could be interpreted as ALL of these (sequences of digits) must be used at least once,

but then adding

"only" carries an implication that the solution is exclusive to those within that given sequence, and not that they all necessarily be utilised.

Also, it is not clear if the solution can include functions such as log_n or power^(1/n)

So, possibly there is a whole raft of solutions, limited only by ones imagination.

Hi Alan and Badinage  

So Badinage, if there is a 'whole raft of solutions' are you going to give us one of them that Alan won't poke holes in ?  LOL

The Question is simple: _+_+_=30

                 5 + 5² + ln1 = 30

Thanks Badinage :))

$${\mathtt{1\,357\,911}} \,{mod}\, {\mathtt{13}}{\mathtt{\,\small\textbf+\,}}{\mathtt{111\,315}} \,{mod}\, {\mathtt{13}}{\mathtt{\,\small\textbf+\,}}{\mathtt{13\,571}} \,{mod}\, {\mathtt{13}} = {\mathtt{30}}$$

BUT only odd numbers up to 15 were supposed to be used