MaxWong

\(3\sqrt{x - 2y}+\dfrac{3}{\sqrt{x-2y}}=10\\ 3\left(\sqrt{x - 2y}\right)^2-10\sqrt{x-2y}+3=0, x \neq 2y\\ \sqrt{x - 2y} = \dfrac{1}{3} \text{ or }\sqrt{x - 2y} = 3\\ x - 2y = \dfrac{1}{9}\text{ or }x - 2y = 9\\ (a - 2)y + b = \dfrac{1}{9}\text{ or }(a - 2)y + b = 9\\ \text{Assuming the above equations are true for all values of }y,\\ (a, b)=\left(2,\dfrac{1}{9}\right)\text{ or }(2,9)\)

\(\begin{array}{cll} &\text{Sum}\\ =&\displaystyle \sum^{\infty}_{n=1} \dfrac{7^{n - 1}}{10^n}\\ =&\dfrac{1}{7}\displaystyle\sum^{\infty}_{n = 1} \left(\dfrac{7}{10}\right)^n\\ \stackrel{\text{sum of GS}}{=} &\dfrac{1}{7}\left(\dfrac{\dfrac{7}{10}}{1-\dfrac{7}{10}}\right)\\ =&\dfrac{1}{3} \end{array}\)

\(a^2 + b^2 = c^2\\ a^2 + b^2 = (b + 1)^2\\ 2b + 1 = a^2\\ \text{Let }a = k^3,\\ k^6 - 1 = 2b\\ k^6 - 1 \equiv 0 \pmod 2\implies k\equiv 1 \pmod 2\\ \text{Exhaust }k.\\ \text{When }k = 1, b = 0\text{(rejected)}\\ \text{When }k = 3, b = 364\\ c = 364 + 1 = \boxed{365}\\ \text{Check:}\\ a^2 + 364^2 = 365^2\\ a = 729 = 9^3\)

From x + y + z = 8 and x - y + z = 4,

\(\color{red}(x+y+z)\color{black}-\color{blue}(x-y+z)\color{black} = \color{red}8\color{black} -\color{blue}4\color{black} = 4\\ 2y = 4\\ y = 2\)
\(x + 2 + z = 8 \\ x + z = 6\)

\(\sqrt{x+z-2}+\sqrt2 = \sqrt x + \sqrt z\\ 2 + \sqrt{2} = \sqrt{x} + \sqrt{z}\\ \text{Let }z = 6-x\\ 2+\sqrt{2} = \sqrt{x} + \sqrt{6 - x}\\ \text{Square both sides,}\\ 6 + 4\sqrt 2 = 6 +2\sqrt{6x-x^2}\\ 2\sqrt 2 = \sqrt{6x - x^2}\\ \text{Square both sides again,}\\ 8 = 6x - x^2\\ x^2 - 6x + 8 = 0\\ (x - 2)(x - 4) = 0\\ x = 2 \text{ or } x = 4\\ \text{When }x = 2,z=4\\ \text{When }x=4,z=2\\ \text{The required ordered triples are }(2,2,4)\text{ and }(4,2,2)\)

I skipped a step. I will show it here.

\(\cdots\\ =\displaystyle \binom72 \cdot 3^7 \cdot \left(\dfrac{5}{3}\right)^2 \cdot x^2\\ \boxed{=\displaystyle \color{red}\binom72\color{black}\cdot \color{blue}3^5 \color{black}\cdot \color{green}5^2\color{black} \cdot x^2}\\ =\color{red} 21 \color{black}\cdot \color{blue}243\color{black}\cdot \color{green}25\color{black}\cdot x^2\)

Infinitely many. Only those with a prime number of occurence of '1's are primes.

I have not been online for quite a while, and I have seldom logged on this site, so I don't know what's happening here.

There are not much difference since I last logged on.

\(\dfrac{x^2 + 1}{x} + \dfrac{x}{x^2 + 1} = \dfrac{29}{10}\\ \text{Let }u = \dfrac{x}{x^2 + 1}.\\ u + \dfrac{1}{u} = \dfrac{29}{10}\\ 10u^2 + 10 - 29u = 0\\ (2u - 5)(5u - 2) = 0\\ u = \dfrac{5}{2} \text{ or } u = \dfrac{2}{5}\\\\ \dfrac{x}{x^2 + 1} \stackrel{(1)}{=} \dfrac{5}{2}\text{ or } \dfrac{x}{x^2 + 1} \stackrel{(2)}{=} \dfrac{2}{5} \\ \text{Consider equation }(1),\\ 5x^2 + 5 = 2x\\ 5x^2 - 2x + 5 = 0\\ x = \dfrac{2 \pm \sqrt{2^2 - 4(5)^2}}{2(5)}\\ x = \dfrac{1\pm\sqrt{-24}}{5}\\ \text{Consider equation }(2),\\ 2x^2 + 2 = 5x\\ 2x^2 - 5x + 2 = 0\\ (2x - 1)(x - 2) = 0\\ \text{All roots are purely real.}\\ \text{The required integer is }-24.\\\)

Oh, yes. Considering only the cases that satisfy the condition is quite troublesome, so I want to make sure that each kid gets one candy at the very beginning, so there will be less constraints to consider.

2)

Consider the pairwise sum of terms.

\(\quad 1-2+3-4+\cdots+2005-2006+2007\\ =(1-2)+(3-4)+\cdots+(2005-2006)+2007\\ =-1+(-1)+\cdots+(-1)+2007\\ \text{There are in total }\dfrac{2006}{2}=1003\text{ pairs that sums to }-1.\\ = -1(1003) + 2007\\ = 1004\)