Rei 7 5 2015-2026

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

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!

I suspect the question is related to knowing whether someone can claim you as a dependent, because otherwise line 5 itself is pretty clear.General answer: if you are under 19, or a full-time student under the age of 24, your parents can probably claim you as a dependent. If you are living with someone to whom you are not married and who is providing you with more than half of your support, that person can probably claim you as a dependent. If you are married and filing jointly, your spouse needs to answer the same questions.Note that whether those individuals actually do claim you as a dependent doesn't matter; the question is whether they can. It is not a choice.

15 + 13 + (11-9) + (7-5) + (1-3) = 30

You don't.I-797, "Notice of Action," is the USCIS form they use to notify you that they received or acted upon something you filed with them. You generally don't fill those at all, you merely receive them.Form I-797: Types and Functions

Depends on the nature of your business and how it is structured.If you own an LLC taxed as a passthrough entity, then you probably will check option b and submit the W7 along with your US non-resident tax return. If your LLC’s income is not subject to US tax, then you will check option a.If the business is a C Corp, then you probably don’t need an ITIN, unless you are receiving taxable compensation from the corporation and then we are back to option b.

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