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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!

TANK CAPACITY = 6 X 50 = 300INITIALLY IT IS (2.3) FREE i.e 200 LITRESREQUIRED LIQUID = 200 LITRES

[math](1\!+\!2\!+\!3\!+\!4)^{5+6-7} \cdot (-8\!+\!9) = 10000[/math]

Every integer can be written in the form of [math]6n,6n+1,6n+2,6n+3,6n+4[/math] or [math]6n+5[/math].It is quite evident that [math]6n[/math] is always divisible by 6, 6n+1 may be prime, [math]6n+2[/math] will always be divisible by 2, [math]6n+3[/math] by 3, and [math]6n+4[/math] by 2. [math]6n+5[/math] may be prime. Our candidates for primality are therefore [math]6n+1[/math] and [math]6n+5[/math]. We can then see that [math]6n+5=6(n+1)-1=6x-1[/math] Therefore, every prime can be written in the form of [math]6n \pm 1[/math]

720 Five-digit numbers will be formed using the digits 1,2, 3,4,5,6. Out of them 360 Five -digit numbers will be even. You can do the math-If we place the 2,4,6 at one’s place the digits formed will be even_ _ _ _ 2Now total no.of ways to arrange 4 numbers out of 5 numbers will be 5! Which is 120Similarly when 4,6 are placed at one’s place, total no.of arrangements are 120 and 120So this gives 360 arrangements.There will be 360 such arrangements which will be even.Thanks for reading my answer i hope you have understand.

It's a classic textbook example of (simple) recursion.(However there's no pruning heuristic to allow early rejection of invalid candidates, since due to all the operations e.g. /, % and √ (and boolean operations like | & ^ and C operations like ~ you can't know if something is way too high/too low/not divisible.So to give a ballpark estimate, if there are B binary operators (neglect unary operators like √ ~ !), and we have to choose one of those to connect the N=9 digits strictly in order with N-1 operators, we must evaluate all (N-1)^B candidates. So for + - * / % that's already 8^5 = 2^15 candidates.If you allow parentheses on two or more adjacent terms there are exponentially more options. I can't enumerate those off the top of my head but let's say there are T terms.And then further if you consider there are U unary operators (e.g. √ ~ ! and maybe unary reductions operators like ^ & |), and if we estimate by assuming that each of the T terms (i.e. either number or parenthesized subexpression) can optionally have one of those U operators applied once, this multiplies the number of candidates by a further factor T*U

This question can be answered using two methods. Let's start with the simplest one.Method 1:The number is three digits, so for them let's take three blanks _ _ _The first blank can be filled using any of the digits from 1–9 because if we use zero to fill the first blank the number becomes of two digits. Hence we have 9 ways to fill the first blank.Now, the second blank can be filled by any of the remaining 10 digits because repetition is allowed and thus the digit selected for the first blank can also be selected. So 10 ways.Similarly 10 ways for the third blank.So total number of combinations become 9 x 10 x 10 = 900Hence the answer is 900 such number can be formed.Method 2:Since the first digit cannot be zero, we have 9C1 ways to select the first digit (one digit selected from a set of nine distinct digits). (9C1 = 9)Now, for the remaining two places we can have zero as well. Hence we have 10C1 ways to select a digit for tens and ones place each. (10C1 = 10)Hence total number of combinations become 9C1 x 10C1 x 10C1 = 9 x 10 x 10 = 900Hence the answer is 900 such numbers can be formed.Hope it helped you! :)