My question had been merged with another one and as a result, I have added the previous answer to the present one. Hopefully this provides a clearer explanation. Just using the numbers given there, it's not possible, because odd + odd = even, even + odd = odd. 30 is an even number, the answer of 3 odd numbers must be odd, it's a contradiction. If what people say is true, then the question is wrongly phrased its any number of operations within those three brackets must lead to 30. Then it becomes a lot easier. Such as 15 + 7 + (7 + 1). That would give 30. But it assumes something that the question does not state explicitly and cannot be done that way. I still stick to my first point, it can't be done within the realm of math and just using three numbers, if not, then the latter is a way to solve it.EDIT:   This question has come up many times, Any odd number can be expressed as the following, Let [math]n, m, p[/math] be an odd number, [math] n = 1 (mod[/math] [math]2), m = 1 (mod[/math] [math]2), p = 1 (mod[/math] [math]2)[/math][math]n+m+p = 1 + 1 + 1 (mod[/math] [math]2)[/math]Let's call [math]n+m+p[/math] as [math]x[/math][math]=> x = 3 (mod[/math] [math]2)[/math]Numbers in modulo n can be added, I'll write a small proof for it below, [math]a = b (mod[/math] [math]n), c = d (mod[/math] [math]n)[/math][math]a+c = b+d (mod[/math] [math]n)[/math]We can rewrite [math]b[/math] and [math]d[/math] in the following way, [math]n | (b - a) => b-a = n*p[/math] (for some integer p) [math]b = a + np[/math][math]b = a + np, d = c + nq[/math][math]b + d = a + np + c + nq[/math][math]b+d = a + c + n(p + q)[/math]Now we have shown that our result is true, moving forward, [math]3 = 1 (mod[/math] [math]2)[/math][math]x = 1 (mod[/math] [math]2)[/math]Therefore the sum of three odd numbers can never be even. It will always be congruent to 1 in mod 2.(This was what I wrote for a merged answer).Modular arithmetic  - Link on modular arithmetic, the basic operations. Modular multiplicative inverse - The multiplicative inverse in modular operations.Congruence relationFermat's little theorem Modular exponentiation - As title suggests.Good luck!