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FAQ

Can you add 5 odd numbers to get 30?
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 do I fill out the educational qualification section of the assistant commandant application form in coast guard (01/2022 batch)?
U should be Bachelor of science hieght166 wt 50 and pass ur exams
How many colleges I can fill out as choices for the All India 15% Quota?
There isn’t any limit as to how many choices you can fill in during the counseling. In the end remember to put those choices at the top which you prefer over others. Seats are allotted based on your preference so choose wisely.
Mathematical Puzzles: What is + + = 30 using 1,3,5,7,9,11,13,15?
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!
How do I get my TDS back if I forgot to fill out my 15H?
simple…file your incometax and claim the deducted TDS (if salary is within no-tax criteria)
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