Substitute Forform 1449PTO Application Number 136 5 3&#39

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!

It is 7,9 + 9,1 + 1 + 3 + 9 = 30Wish you can find the 7,9 and 9,1 in the list of1,3,5,   7,9    ,11,13,151,3,5,7,      9,1    1,13,15

“How many ways are there to make 5 out of just the numbers 1, 2, 3, 4, 6, 7 by adding, subtracting, multiplying and dividing?”An infinite number of ways, since you didn’t say how many times I was allowed to repeat a digit, or a limit to how long an expression I could make.First, pick any “5”; let’s use “4+1”.Now, let’s create a bunch of “zero”-patterns:(4+2–6), (3+4–7), (2+1–3), or even longer ones like (7+6–4–3–3–2–1)Next, add as many zero patterns as you like, in whatever order you like, until you get tired of doing that. Then, since you tire so easily, write (or have someone write for you) a computer program to keep appending the next random choice of zero pattern to your ever-growing expression line.You can keep doing this until the universe ends, so there is no limit to the number of ways.

void print(int n)