MaxWong

Hint: When x = 1, y = A + B + C. Try to find the y-coordinate of the point on the graph with x-coordinate = 1.

It is not possible for a function to have different values for the same input. f(3) cannot be 0 and 28 at the same time.

Plotting it in a graphical calculator gives the range as \((-\infty, +\infty)\).

If n/21 is reducible, then either n is a multiple of 3 or n is a multiple of 7.

There are floor(100/3) = 33 multiples of 3 from 1 to 100, and there are floor(100/7) = 14 multiples of 7 from 1 to 100.

But we have double counted floor(100/21) = 4 multiples of 21, so the total number of integers n with 1 <= n <= 100 such that n/21 is reducible is 33 + 14 - 4 = 43.

\(\dfrac{x^2 + 5x + a}{x^2 - 10x - 24} = \dfrac{x^2 + 5x + a}{(x - 12)(x + 2)}\)

If the fraction is equal to the quotient of two linear functions, then either (x - 12) or (x + 2) divides x^2 + 5x + a.

By factor theorem, 

\(12^2 + 5\cdot 12 + a = 0\text{ or }(-2)^2 + 5(-2) + a = 0\\ a = -204 \text{ or }a = 6\)

Therefore \(S = \{-204, 6\}\). Sum of elements of S is -198.

Remarks: The statement of factor theorem is as follows.

Let \(p(x)\) be a polynomial such that \((x - a)\) divides \(p(x)\). Then \(p(a) = 0\).

Let AR = 2k, RC = 3k, AP = p, PQ = 2p, QB = p.

Then 

\(\begin{array}{rcl} [QBC] &=& [ABC] - [ACQ]\\ &=& \dfrac12 (5k)(4p) \sin \angle A - \dfrac12 (5k)(3p) \sin \angle A\\ &=& \dfrac{5kp}2 \sin \angle A\\ [CRQ] &=& [ACQ] - [AQR]\\ &=& \dfrac12 (5k)(3p)\sin \angle A - \dfrac12 (2k)(3p)\sin \angle A\\ &=& \dfrac{9kp}2\sin \angle A \end{array}\)

Therefore, 

\(\dfrac{[QBC]}{[CRQ]} = \dfrac{\dfrac{5kp}2\sin \angle A}{\dfrac{9kp}2\sin \angle A} = \dfrac5{9}\)

See this answer: https://web2.0calc.com/questions/number-theory_16563#r1

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