MaxWong

\(3Ax-24=5x-9+x\\ (3A-6)x=15\\ x = \dfrac{5}{A-2}\\ \text{When }A = 2, \text{there are no solutions.}\)

\(z^2+z+1 = 0\\ (z-1)(z^2+z+1)=0\\ z^3 = 1, z \neq 1\\ z = -\dfrac{1}{2} + \dfrac{i\sqrt3}{2}\\ z^2 = -\dfrac{1}{2} - \dfrac{i\sqrt3}{2}\\\)

\(z^{49} + z^{50} + z^{51} + z^{52} + z^{53}\\ =z + z^2 + 1 + z + z^2\\ =1+2(-\dfrac{1}{2} + \dfrac{i\sqrt3}{2} + -\dfrac{1}{2} - \dfrac{i\sqrt3}{2})\\ =1 + 2(-1)\\ = -1\)

z^3 = x^3 - iy^3 + 3xyi(x+yi) = x^3 - 3xy^2 + 3x^2yi-iy^3

Re(z^3) = x(x^2-3y^2)

x(x^2-3y^2) = 74. 

Considering all possible values of x,

(x, y) = (1, 5).

c = Im(z^3) = 3(5) - 5^3 = -110.

Let d be the diameter.

\(d^2 = (-1-3)^2 + (-2-2)^2 = 32\)

Formula of area of circle: \(A = \dfrac{\pi d^2}{4} = \dfrac{32\pi}{4} = 8\pi\)

Finding the absolute differences of every possible 2-permutation of these 3 integers and writing out its prime factorisation:

|1457 - 368| = 1089 = 3^2 * 11^2

|368 - 1754| = 1386 = 2 * 3^2 * 7 * 11

|1754 - 1457| = 297 = 3^3 * 11

Finding the G.C.D of these absolute differeces:

GCD(3^2 * 11^2, 2*3^2*7*11, 3^3*11) = 3^2 * 11 = 99

Therefore 99 is the largest positive integer such that 1457, 368, and 1754 leave the same remainder when divided by it.

(Why does this work? The proof is trivial and is left for the reader as an exercise. )

Given that \(-3 \le 7x+2y\le 3\text{ and }-4\le y -x \le 4\), what is the maximum possible value of x + y?

\(-10\le \dfrac{5y}{2} - \dfrac{5x}{2} \le 10\)

\(-13 \le \dfrac{9x}{2} + \dfrac{9y}{2} \le 13\)

\(\dfrac{-26}{9}\le x+y \le \dfrac{26}{9}\)

Maximum possible value is 26/9

\(x^2+2xy+3y^2 -6x-2y\\ =(x+y)^2 + 2y^2-2y-6x\\ =(x+y)^2 + 2y(y-1) - 6x\\ \boxed{\text{Minimum occurs when x = -y.}}\\ = 2y^2-2y+6y\\ = 2(y^2+2y)\\ =2((y+1)^2-1)\\ =2(y+1)^2 - 2\\ \text{Minimum value} = -2.\)

Let P be the value of the required expression.

\(P = \dfrac{cd+bd+bc+ad+ac+ab}{abcd} = \dfrac{\dfrac{9}{1}}{\dfrac{4}{1}} = \dfrac{9}{4}\)

Q: Find all 6-digit multiples of 22 of the form 5d522e where d and e are digits. What's the maximum value of d?

5d522e is divisible by 22 <=> 5d522e is divisible by 2 and also divisible by 11.

e = {0,2,4,6,8}.

5 + 5 + 2 - d - 2 - e is divisible by 11.

\(14-d-e\equiv 0 \pmod {11}\\ d+e\equiv 3 \pmod{11}\\ (d,e) = (3,0), (1,2),(8,6),(6,8)\)

The required multiples of 22 are 535220, 515222, 585226, 565228.

Maximum value of d is 8.

\(\sin\left(120^{\circ}\right)\\ = 2\sin\left(60^{\circ}\right)\cos\left(60^{\circ}\right)\\ =2\left(\dfrac{\sqrt3}{2}\right)\left(\dfrac{1}{2}\right)\\ =\dfrac{\sqrt3}2\)   

 \(\tan\left(60^{\circ}\right)\\ =\dfrac{2\tan\left(30^{\circ}\right)}{1-\tan^2\left(30^{\circ}\right)}\\ =\dfrac{\dfrac{2}{\sqrt3}}{1-\dfrac{1}{3}}\\ =\dfrac{\dfrac{2}{\sqrt3}}{\dfrac{2}{3}}\\ =\dfrac{3}{\sqrt3}\\ =\sqrt3\)

\(\cos\left(\dfrac{4\pi}{3}\right)\\ = 2\cos^2\left(\dfrac{2\pi}3{}\right) - 1\\ = 2\left(2\cos^2\left(\dfrac{\pi}{3}\right)-1\right)^2-1\\ =2\left(2\left(\dfrac{1}{2}\right)^2-1\right)^2 - 1\\ =2\left(-\dfrac{1}{2}\right)^2 - 1\\ =-\dfrac{1}{2}\)

\(\sin\left(\dfrac{5\pi}{3}\right)\\ = 2\sin\left(\dfrac{5\pi}{6}\right)\cos\left(\dfrac{5\pi}{6}\right)\\ = 2\left(\dfrac{1}{2}\right)\left(\dfrac{-\sqrt3}{2}\right)\\ =\dfrac{-\sqrt3}2\)