Kromy

See my previous answer to your $a_{12}$ question. It should be a similar (if not the exact same) process with Cauchy Schwarz. 

Edit: my previous answer got large ai language modeled. 

Here is how to start with cauchy: $x_i = \sqrt{i} \implies \sum_{i=1} ^{20}{x_i} = \sqrt{1} + \sqrt{2} + ... + \sqrt{20} \implies \sum_{i=1} ^{20}{x_i^2} = 1 + 2 + ... + 20$. Cauchy says sum of x^2 * sum of y^2 >= sum of (xy)^2. 

Please think this problem through, it is a basic problem that shouldn't require any help. Here are some hints: 

Note that each number is 6 plus the last number. So can we manipulate the sum to be divisible by six? That way we can apply the formula $1 + 2 + ... + n = \frac{n(n-1)}{2}$? 

Thank you for putting it in latex so its readable :) 

The basic idea when we see the word minimize is to think of inequalities. I suspect you got this problem from a book about inequalities. We also see squares, and that gives us the idea to use Cauchy Schwarz inequality. As a recap, it states $\left| \sum_{i=1}^na_ib_i \right|^2 \le \left(\sum_{i=1}^{n}|a_i^2| \right) \left( \sum_{i=1}^n |b_i^2| \right)$. See here: https://artofproblemsolving.com/wiki/index.php/Cauchy-Schwarz_Inequality?srsltid=AfmBOore-GDD9oAtWZDDSJ-URH43dfQd2atn26PRaLxAr01bGo2-r183 (aops wiki page). 

This indeed matches with minimizing squares. Now, the question becomes, what do we let $a_i$ and $b_i$ be? [Looking back on it, we shouldn't use $a_i$to avoid confusion, but oh well... our "$a_i$" here is refering to the Cauchy Schwarz inequality, and not to the problem.] Well, let's look at our coefficients. we have 1, 2, 3, ... , 20. Since we are squaring things in our formula, we probably want something in the form $\sqrt{1}, \sqrt{2}, \sqrt{3}, ... , \sqrt{20}$. So if we have$a_i = \sqrt{i} \implies \sum_{i=1} ^{20}{a_i} = \sqrt{1} + \sqrt{2} + ... + \sqrt{20} \implies \sum_{i=1} ^{20}{a_i^2} = 1 + 2 + ... + 20$. What about $b_i$? We have $a_i$ in the form of some square roots, but when we want to minimize $\left| \sum_{i=1}^na_ib_i \right|^2$, we have $A_i$ terms and whole numbers (here, $A_i$ stands for the real numbers given in the problem). So, we want a $b_i$ such that when it is multiplied by $a_i$, we get nice coefficients. It becomes obvious (at least to me), that $\sum_{i=1} ^{20}{b_i} = \sqrt{1} A_1 + \sqrt{2}A_2 + \sqrt{3}A_3 ... + \sqrt{20} A_{20} \implies \sum_{i=1} ^{20}{b_i ^ 2} = 1 A_1^2 + 2A_2^2 + 3A_3^2 ... + 20 A_{20}^2.$ Then, $\sum_{i=1} ^{20}{(a_i b_i)^2} = 1A_1 + 2A_2 + 3A_3 ... + 20 A_{20}$. 

Finally, putting it all together, we have $1A_1 + 2A_2 + 3A_3 + ... 20A_{20} \le (1 + 2 + 3 + ... + 20) (1A_1^2 + 2A_2^2 + 3A_3 ^ 2 + ... + 20 A_{20}^2)$. From the problem, we have $1A_1 + 2A_2 + 3A_3 + ... 20A_{20} = 1$, and we know the sum of $(1 + 2 + 3 + ... + 20) = 210$(You can think of it as grouping: 1+20 = 21, 2+19 = 21, 3+18 = 21. We have 10 total pairs, so 21x10 = 210. This is the derivation of the formula: $1 + 2 + ... + n = \frac{n(n-1)}{2}$). ]

Plugging these values in, we have $\frac{1}{210} \le (1A_1^2 + 2A_2^2 + 3A_3 ^ 2 + ... + 20 A_{20}^2)$. 

Now, to minimize this, we need equality to hold. The equality condition is as follows: 

We need some $x$ such that $x a_i = b_i$, or $x \sqrt{i} = \sqrt{i} A_i$, so $x = A_i$ for all $i \in {(1, 2, 3, ..., 20)}$[The "e" notation is reads "in", so for all i "in" (1, 2, ..., 20).] 

From the original problem, we have $1A_1 + 2A_2 + 3A_3 + ... 20A_{20} = 1$, so plugging in $x = A_i$ gives $1 + 2x + 3x + ... 20 x = 1 \implies x(1+2+3 + ... + 20) = 1, \text{so } x = \frac{1}{120}$. So, $a_{12} = 12 \cdot \frac{1}{120} = \frac{1}{10}$. 

Note: I may have made some mistakes on the way, so make sure to read the solution, understand everything, and see what mistakes I made... Because just getting the answer without learning doesn't help you  

For convienence, why don't we let a = x - 3. Our equation becomes a^4 + (a-2)^4 = -8 + 6(a)(a-2)^3 - 11(a)^3 (a-2). Expanding and simplfiying, we have 7 a^4 + 6 a^3 - 48 a^2 + 16 a + 24= 0. The best way to do this problem now is plug in values using the Rational Root Theorem until you have one, then factor it out.

However, from here, there is no good way to proceed, as the rational root theorem test values all fail. I would like to assume there is a typo with one of the numbers in your equation, because wolframalpha is giving me x≈-0.15270, x≈2.4454, x≈2.4454, x≈4.6898. 

Good luck, the general approach should be there, just double check your numbers! 

I'm assuming your "bc" is a typo and the polynomial is p(x) = x^3+ax^2+bx+c.

We know that the y intercept occurs at (0, 0), which implies that C = 0. By Vieta's theorem, is $C= {-r \cdot s \cdot t}$, where r, s, t are the roots of p(x). So at the minimum, either r, s, or t is 0. Without loss of generality, let's assume r = 0. 

We know rst = (r + s + t)/3 = 1 - (r + s + t) + (rs + st + rt) - (rst), by Vieta's theorem. Plugging in r = 0, we have $0 = \frac{s+t}{3} = 1 - (s+t) + (st)$. Then, we can seperate this big equation into one equation (by ignoring the 3rd equality): $s+t = 0 \Rightarrow s = -t$ . Now, we plug in s = -t to the 3rd expression to get $1 - 0 - t^2 = 0 \Rightarrow t^2 = 1, t = -1, 1$. Now, depending on what t is, s is either -1 or 1. So let's just let t = 1 (it won't affect the problem's final answer anyways) and then s = -1. So then b = (rs + st + rt) = (st) = -1. 

So, our final answer is -1. This problem is fairly easy, and is a simple application of Vieta's theorem, which is absolutly necessary to know for polynomial Algebra questions, and comes up often in math competitions like the AMCs :) 

Other than that, some "tricks" (not really) to note that is if you want the y-intercept, x is 0 and thus the y-intercept will always be C. If you want the sum of the coefficients, this is the equivalent to x = 1 (applies vice versa!). 

psss i was too lazy to write out all the latex, writing latex on this website is hard 

What do you mean "answer?" You have given us two variables, with only one equation. There are infinite "solutions"... If you want to assume that x and y are integers and make it a diophantine equation, that is a different matter. 

Here is y in terms of x: 

y = (75 + 72x)/86

(n+1)^2/(n+2). Expanding, we have (n^2 + 2n + 1)/(n+2). By synthetic division, we have n + 1/(n+2). Now we want a positive integer n such that 1/(n+2) is an integer. This obviously cannot happen, we would need -1 to make 1/(n+2) an integer. 

(Going off topic here), theoritically, if you take the limit as n approaches infinity, 1/(n+2) is 0 and you get infinity. I guess it would be divergent either way (awww). 

Final answer is cannot happen, there isnt an n that can satisfy the given conditions. 

Let N = 1000 a + 100 b + 10 c + d. Then, we want to have the relationship: (1000 a + 100 b + 10 c + d)/24 = 100 b + 10 c + d. 

Let M = 100 b + 10 c + d. It follows that N = 1000 a + M. We also know M = N/24. Thus, we have M = (1000 a + M)/24. Solving for M, we have M = 1000a/23. Now, no matter what integer from 1-9 a takes on, M will never be an integer. Therefore there are 0 positive integers N that has the property stated.