20250922 Senior Varsity Training

Problem 1

Find the number of integer values of $k$ in the closed interval $[- 500, 500]$ for which the equation

lo g (k x) = 2 lo g (x + 2)

has exactly one real solution.

Solution

$lo g (k x) = 2 lo g (x + 2) ⟺ x > - 2 \land lo g (k x) = lo g ((x + 2)^{2}) ⟺ x > - 2 \land k x > 0 \land k x = (x + 2)^{2} ⟺ x > - 2 \land k x > 0 \land x^{2} + (4 - k) x + 4 = 0$
Let $f (x) = x^{2} + (4 - k) x + 4 = (x - x_{1}) (x - x_{2})$ , where $x_{1} = \frac{k - 4 - k ( k - 8 )}{2}, x_{2} = \frac{k - 4 + k ( k - 8 )}{2}$ , then
$lo g (k x) = 2 lo g (x + 2) ⟺ x > - 2 \land k x > 0 \land f (x) = 0 ⟺ x > - 2 \land k x > 0 \land (x = x_{1} \lor x = x_{2}) ⟹ k < 0 \lor k \geq 8$
Applying Vieta’s Formulas yields
$x_{1} + x_{2} = k - 4 \land x_{1} x_{2} = 4$
If $k < 0$ ,
$x_{1} + x_{2} = k - 4 \land x_{1} x_{2} = 4 ⟹ x_{1} + x_{2} < 0 \land x_{1} x_{2} > 0 ⟺ x_{1} < 0 \land x_{2} < 0$
Since $f^{''} (x) = 2 > 0 \land f (- 2) = 2 k < 0$ , it follows that
$x_{1} < - 2 \land x_{2} > - 2$
Therefore,
$lo g (k x) = 2 lo g (x + 2) ⟺ x = x_{2}$
Therefore, if $k < 0$ , $lo g (k x) = 2 lo g (x + 2)$ always has exactly one real solution.

If $k \geq 8$ ,
$x_{1} + x_{2} = k - 4 \land x_{1} x_{2} = 4 ⟹ x_{1} + x_{2} > 0 \land x_{1} x_{2} > 0 ⟺ x_{1} > 0 \land x_{2} > 0$
Therefore,
$lo g (k x) = 2 lo g (x + 2) ⟺ x = x_{1} \lor x = x_{2}$
Therefore, if $k \geq 8$ , $lo g (k x) = 2 lo g (x + 2)$ has exactly one real solution iff
$x_{1} = x_{2} ⟺ k (k - 8) = 0 ⟺ k = 8$
Therefore, $lo g (k x) = 2 lo g (x + 2)$ has exactly one real solution iff
$k < 0 \lor k = 8$
There are $501$ integer values of $k$ in $[- 500, 500]$ such that $k < 0 \lor k = 8$ .

Problem 2

Real numbers $x$ and $y$ with $x, y > 1$ satisfy $lo g_{x} (y^{x}) = lo g_{y} (x^{4 y}) = 10$ . What is the value of $x y$ ?

Solution

$lo g_{x} (y^{x}) lo g_{y} (x^{4 y}) = x lo g_{x} (y) \cdot 4 y lo g_{y} (x) = 4 x y$
Therefore,
$4 x y = 100$ $4 x y = 100 ⟺ x y = 25$

Problem 3

Let $x$ , $y$ and $z$ be positive real numbers that satisfy the following system of equations:

lo g_{2} (\frac{x}{yz}) = \frac{1}{2}, lo g_{2} (\frac{y}{x z}) = \frac{1}{3}, lo g_{2} (\frac{z}{x y}) = \frac{1}{4}

Then the value of $lo g_{2} (x^{4} y^{3} z^{2})$ is $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution

$lo g_{2} (\frac{x}{yz}) = \frac{1}{2} \land lo g_{2} (\frac{y}{x z}) = \frac{1}{3} \land lo g_{2} (\frac{z}{x y}) = \frac{1}{4} ⟺ lo g_{2} (x) - lo g_{2} (y) - lo g_{2} (z) = \frac{1}{2} \land lo g_{2} (y) - lo g_{2} (x) - lo g_{2} (z) = \frac{1}{3} \land lo g_{2} (z) - lo g_{2} (x) - lo g_{2} (y) = \frac{1}{4} ⟺ lo g_{2} (x) = - \frac{7}{24} \land lo g_{2} (y) = - \frac{3}{8} \land lo g_{2} (z) = - \frac{5}{12}$
Therefore,
$lo g_{2} (x^{4} y^{3} z^{2}) = ∣ 4 lo g_{2} (x) + 3 lo g_{2} (y) + 2 lo g (z) ∣ = \frac{25}{8}$
Therefore, $m + n = 33$ .

Problem 4

Let $a > 1$ and $x > 1$ satisfy

lo g_{a} (lo g_{a} (lo g_{a} (2)) + lo g_{a} (2) - 128) = 128 \land lo g_{a} (lo g_{a} (x)) = 256

Find the reminder when $x$ is divided by $1000$ .

Solution

$lo g_{a} (lo g_{a} (lo g_{a} (2)) + lo g_{a} (24) - 128) = 128 ⟺ lo g_{a} (24 lo g_{a} (2)) = 128 + a^{128} ⟺ lo g_{a} (2^{24}) = a^{128} a^{a^{128}} ⟺ 2^{24} = (a^{a^{128}})^{a^{a^{128}}} ⟺ a^{a^{128}} = 2^{3} ⟺ a = 2^{\frac{3}{64}}$
Therefore,
$lo g_{a} (lo g_{a} (x)) = 256 ⟺ x = a^{a^{256}} ⟺ x = 2^{192} ⟹ x \equiv 896 (mod 1000)$

Problem 5

Let $x$ , $y$ and $z$ be real numbers satisfying the system

lo g_{2} (x yz - 3 + lo g_{5} (x)) = 5 lo g_{3} (x yz - 3 + lo g_{5} (y)) = 4 lo g_{4} (x yz - 3 + lo g_{5} (z)) = 4

Find the value of $∣ lo g_{5} (x) ∣ + ∣ lo g_{5} (y) ∣ + ∣ lo g_{5} (z) ∣$ .

Solution

$lo g_{2} (x yz - 3 + lo g_{5} (x)) = 5 \land lo g_{3} (x yz - 3 + lo g_{5} (y)) = 4 \land lo g_{4} (x yz - 3 + lo g_{5} (z)) = 4 ⟺ x yz + lo g_{5} (x) = 35 \land x yz + lo g_{5} (y) = 84 \land x yz + lo g_{5} (z) = 259 ⟹ 3 x yz + lo g_{5} (x yz) = 378 ⟺ x yz = 125$
Therefore,
$x yz + lo g_{5} (x) = 35 ⟺ lo g_{5} (x) = - 90$ $x yz + lo g_{5} (y) = 84 ⟺ lo g_{5} (y) = - 41$ $x yz + lo g_{5} (z) = 259 ⟺ lo g_{5} (z) = 134$
Therefore,
$∣ lo g_{5} (x) ∣ + ∣ lo g_{5} (y) ∣ + ∣ lo g_{5} (z) ∣ = 265$

Problem 6

Quadratic polynomials $P (x)$ and $Q (x)$ have leading coefficients of $2$ and $- 2$ , respectively. The graphs of both polynomials pass through the two points $(16, 54)$ and $(20, 53)$ . Find $P (0) + Q (0)$ .

Solution

Let $P (x) = 2 x^{2} + b_{P} x + c_{P}$ , then
$P (16) = 54 \land P (20) = 53 ⟺ 512 + 16 b_{P} + c_{P} = 54 \land 800 + 20 b_{P} + c_{P} = 53 ⟺ b_{P} = - \frac{289}{4} \land c_{P} = 698$
Therefore,
$P (x) = 2 x^{2} - \frac{289}{4} x + 698$
Let $Q (x) = - 2 x^{2} + b_{Q} x + c_{Q}$ , then
$Q (16) = 54 \land Q (20) = 53 ⟺ - 512 + 16 b_{Q} + c_{Q} = 54 \land - 800 + 20 b_{Q} + c_{Q} = 53 ⟺ b_{Q} = \frac{287}{4} \land c_{Q} = - 582$
Therefore,
$Q (x) = - 2 x^{2} + \frac{287}{4} x - 582$
Therefore,
$P (0) + Q (0) = 698 - 582 = 116$

Problem 7

For certain real numbers $a$ , $b$ , and $c$ , the polynomial

g (x) = x^{3} + a x^{2} + x + 10

has three distinct roots, and each root of $g (x)$ is also a root of the polynomial

f (x) = x^{4} + x^{3} + b x^{2} + 100 x + c

Solution

Let $x_{0}$ be the additional root of $f (x)$ , then
$f (x) = (x - x_{0}) g (x) = (x - x_{0}) (x^{3} + a x^{2} + x + 10) = x^{4} + (a - x_{0}) x^{3} + (1 - a x_{0}) x^{2} + (10 - x_{0}) x - 10 x_{0}$
Therefore,
$a - x_{0} = 1 \land 1 - a x_{0} = b \land 10 - x_{0} = 100 \land - 10 x_{0} = c$ $a - x_{0} = 1 \land 1 - a x_{0} = b \land 10 - x_{0} = 100 \land - 10 x_{0} = c ⟺ x_{0} = - 90 \land a = - 89 \land b = - 8009 \land c = 900$
Therefore,
$f (x) = x^{4} + x^{3} - 8009 x^{2} + 100 x + 900$
Therefore, $f (1) = - 7007$ .

Problem 8

There are nonzero integers $a$ , $b$ , $r$ and $s$ such that the complex number $r + s i$ is a zero of the polynomial $P (x) = x^{3} - a x^{2} + b x - 65$ . For each possible combination of $a$ and $b$ , let $ρ_{a, b}$ be the sum of the zeros of $P (x)$ . Find the sum of the $ρ_{a, b}$ ‘s for all possible combinations of $a$ and $b$ .

Solution

It is easy to prove that the problem is equivalent to finding the sum of the sum of zeros of all $P (x) \in R [x]$ such that
$\exists r, s \in Z ∖ {0}, P (r + s i) = 0 \land \exists a, b \in Z ∖ {0}, P (x) = x^{3} - a x^{2} + b x - 65$
Since $P (x) \in R [x]$ , it follows that
$P (r + s i) = 0 ⟺ P (r - s i) = 0$
Therefore,
$\exists r, s \in Z ∖ {0}, P (r + s i) = 0 ⟺ \exists r, s \in Z ∖ {0}, \exists t \in R, P (x) = (x - r + s i) (x - r - s i) (x - t) ⟺ \exists r, s \in Z ∖ {0}, \exists t \in R, P (x) = x^{3} - (2 r + t) x^{2} + (r^{2} + s^{2} + 2 r t) x - (r^{2} + s^{2}) t$
Therefore,
$\exists r, s \in Z ∖ {0}, P (r + s i) = 0 \land \exists a, b \in Z ∖ {0}, P (x) = x^{3} - a x^{2} + b x - 65 ⟺ \exists r, s \in Z ∖ {0}, \exists t \in R, P (x) = x^{3} - (2 r + t) x^{2} + (r^{2} + s^{2} + 2 r t) x - (r^{2} + s^{2}) t \land 2 r + t \in Z ∖ {0} \land r^{2} + s^{2} + 2 r t \in Z ∖ {0} \land (r^{2} + s^{2}) t = 65 ⟺ \exists r, s \in Z ∖ {0}, \exists t \in Z_{+}, P (x) = x^{3} - (2 r + t) x^{2} + (r^{2} + s^{2} + 2 r t) x - (r^{2} + s^{2}) t \land (r^{2} + s^{2}) t = 65 ⟺ \exists ⟨ r, s, t ⟩ \in {⟨ 1, 2, 13 ⟩, ⟨ 2, 1, 13 ⟩, ⟨ - 1, - 2, 13 ⟩, ⟨ - 2, - 1, 13 ⟩, ⟨ 2, 3, 5 ⟩, ⟨ 3, 2, 5 ⟩, ⟨ - 2, - 3, 5 ⟩, ⟨ - 3, - 2, 5 ⟩, ⟨ 1, 8, 1 ⟩, ⟨ 8, 1, 1 ⟩, ⟨ - 1, - 8, 1 ⟩, ⟨ - 8, - 1, 1 ⟩ ⟨ 4, 7, 1 ⟩, ⟨ 7, 4, 1 ⟩, ⟨ - 4, - 7, 1 ⟩, ⟨ - 7, - 4, 1 ⟩}, P (x) = (x - r + s i) (x - r - s i) (x - t)$
Therefore, the sum of the sum of zeros of $P (x)$ is $80$ .

Problem 9

Find all polynomials $f (x)$ with real coefficients such that $f (x^{2}) = f (x)^{2}$ .

Solution

$f (x^{2}) = f (x)^{2} ⟹ \forall x \in C, f (x^{2}) = 0 \Leftrightarrow f (x)^{2} = 0 ⟺ \forall x \in C, f (x^{2}) = 0 \Leftrightarrow f (x) = 0$
Therefore, if $\exists x \in C ∖ {- 1, 0, 1}, f (x) = 0$ , infinitely many distinct numbers
$x, x^{2}, x^{4}, x^{8}, \dots$
are all zeros of $f (x)$ . If $\exists x \in {0, 1}, f (x) = 0$ , infinitely many distinct numbers
$e^{iπ}, e^{\frac{iπ}{2}}, e^{\frac{iπ}{4}}, e^{\frac{iπ}{8}}, \dots$
are all zeros of $f (x)$ . Therefore,
$f (x^{2}) = f (x)^{2} ⟹ f (x) = 0 \lor \forall x \in C ∖ {0}, f (x) \neq = 0 ⟺ \exists a \in R, \exists n \in N, f (x) = a x^{n}$
Therefore,
$f (x^{2}) = f (x)^{2} ⟺ \exists a \in R, \exists n \in N, f (x) = a x^{n} \land a x^{2 n} = a^{2} x^{2 n} ⟺ \exists a \in R, \exists n \in N, f (x) = a x^{n} \land (a = 0 \lor a = 1) ⟺ f (x) = 0 \lor \exists n \in N, f (x) = x^{n}$

Problem 10

There exist two triples of real numbers $⟨ a, b, c ⟩$ such that $a - \frac{1}{b}$ , $b - \frac{1}{c}$ , and $c - \frac{1}{a}$ are the roots to the cubic equation $x^{3} - 5 x^{2} - 15 x + 3$ listed in increasing order. Denote those $⟨ a_{1}, b_{1}, c_{1} ⟩$ and $⟨ a_{2}, b_{2}, c_{2} ⟩$ . If $a_{1}$ , $b_{1}$ and $c_{1}$ are the roots to monic cubic polynomial $f$ and $a_{2}$ , $b_{2}$ and $c_{2}$ are the roots to monic cubic polynomial $g$ , find $f (0)^{3} + g (0)^{3}$ .

Solution

Applying Vieta’s Formulas yields
$(a - \frac{1}{b}) (b - \frac{1}{c}) (c - \frac{1}{a}) = - 3$ $(a - \frac{1}{b}) (b - \frac{1}{c}) (c - \frac{1}{a}) = - 3 ⟺ ab c - a - b - c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - \frac{1}{ab c} = - 3$
Applying Vieta’s Formulas yields
$a - \frac{1}{b} + b - \frac{1}{c} + c - \frac{1}{a} = 5$
Therefore,
$ab c - \frac{1}{ab c} = 2$ $ab c - \frac{1}{ab c} = 2 ⟺ (ab c)^{2} - 2 ab c - 1 = 0 ⟺ ab c = 1 - 2 \lor ab c = 1 + 2$
Applying Vieta’s Formulas yields
$f (0) = - a_{1} b_{1} c_{1}, g (0) = - a_{2} b_{2} c_{2}$
It is easy to prove that $f (0) \neq = g (0)$ , therefore
$f (0)^{3} + g (0)^{3} = (- (1 - 2))^{3} + (- (1 + 2))^{3} = - 14$

Problem 11

Find all possible values of $a$ such that the roots of polynomial $x^{3} - 6 x^{2} + a x + a$ (denote by $x_{1}, x_{2}, x_{3}$ ) satisfy that $(x_{1} - 3)^{3} + (x_{2} - 3)^{3} + (x_{3} - 3)^{3} = 0$ .

Solution

Applying Newton’s Identities yields
$x_{1} + x_{2} + x_{3} = 6$ $x_{1}^{2} + x_{2}^{2} + x_{3}^{2} = 6 (x_{1} + x_{2} + x_{3}) - 2 a = 36 - 2 a$ $x_{1}^{3} + x_{2}^{3} + x_{3}^{3} = 6 (x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) - a (x_{1} + x_{2} + x_{3}) - 3 a = 216 - 21 a$
Therefore,
$(x_{1} - 3)^{3} + (x_{2} - 3)^{3} + (x_{3} - 3)^{3} = x_{1}^{3} + x_{2}^{3} + x_{3}^{3} - 9 (x_{1}^{2} + x_{2}^{2} + x_{3}^{2}) + 27 (x_{1} + x_{2} + x_{3}) - 81 = - 3 a - 27$
Therefore,
$(x_{1} - 3)^{3} + (x_{2} - 3)^{3} + (x_{3} - 3)^{3} = 0 ⟺ - 3 a - 27 = 0 ⟺ a = - 9$

Problem 12

If $r$ , $s$ , $t$ and $u$ denote the roots of the polynomial $f (x) = x^{4} + 3 x^{3} + 3 x + 2$ , evaluate $\frac{1}{r ^{2}} + \frac{1}{s ^{2}} + \frac{1}{t ^{2}} + \frac{1}{u ^{2}}$ .

Solution

Let
$g (x) = x^{n} f (x^{- 1}) = 2 x^{4} + 3 x^{3} + 3 x + 1$
then, since
$g (x) = x^{n} f (x^{- 1}) = (r x - 1) (s x - 1) (t x - 1) (ux - 1)$
$g (x)$ is a polynomial with roots $\frac{1}{r}$ , $\frac{1}{s}$ , $\frac{1}{t}$ and $\frac{1}{u}$ .

Applying Newton’s Identities yields
$\frac{1}{r} + \frac{1}{s} + \frac{1}{t} + \frac{1}{u} = - \frac{3}{2}$ $\frac{1}{r ^{2}} + \frac{1}{s ^{2}} + \frac{1}{t ^{2}} + \frac{1}{u ^{2}} = - \frac{3}{2} (\frac{1}{r} + \frac{1}{s} + \frac{1}{t} + \frac{1}{u}) = \frac{9}{4}$

Problem 13

Find all polynomials of degree less than $4$ satisfying that $f (0) = 1, f (1) = 1, f (2) = 5, f (3) = 11$ .

Solution

It is easy to prove the problem is equivalent to finding all polynomials $f (x)$ such that
$\exists a, b, c, d \in Z, f (x) = a x^{3} + b x^{2} + c x + d \land f (0) = 1 \land f (1) = 1 \land f (2) = 5 \land f (3) = 11$ $\exists a, b, c, d \in Z, f (x) = a x^{3} + b x^{2} + c x + d \land f (0) = 1 \land f (1) = 1 \land f (2) = 5 \land f (3) = 11 ⟺ \exists a, b, c, d \in Z, f (x) = a x^{3} + b x^{2} + c x + d \land d = 1 \land a + b + c + d = 1 \land 8 a + 4 b + 2 c + d = 5 \land 27 a + 9 b + 3 c + d = 11 ⟺ \exists a, b, c, d \in Z, f (x) = a x^{3} + b x^{2} + c x + d \land a = - \frac{1}{3} \land b = 3 \land c = - \frac{8}{3} \land d = 1 ⟺ f (x) = - \frac{1}{3} x^{3} + 3 x^{2} - \frac{8}{3} x + 1$

Problem 14

What is the sum of the roots of $z^{12} = 64$ that have a positive real part?

Solution

The roots of $z^{12} = 64$ are
$z_{k} = 2 (cos (\frac{kπ}{6}) + i sin (\frac{kπ}{6})), k = 0, 1, \dots, 11$
Since
$cos (\frac{kπ}{6}) > 0 ⟺ k \in {0, 1, 2, 10, 11}$
It follows that the roots with a positive real part are $z_{0}, z_{1}, z_{2}, z_{10}, z_{11}$ .
$k \in {0, 1, 2, 10, 11} \sum z_{k} = 2 k \in {0, 1, 2, 10, 11} \sum cos (\frac{kπ}{6}) + 2 i k \in {0, 1, 2, 10, 11} \sum sin (\frac{kπ}{6}) = 2 (cos (0) + 2 cos (\frac{π}{6}) + 2 cos (\frac{π}{3})) = 22 + 6$

Problem 15

The equation $x^{2} - 2 a x + a^{2} - 4 a = 0, a \in R$ has at least one root with magitude $3$ . Find all possible values of $a$ .

Solution

Let $f (x) = x^{2} - 2 a x + a^{2} - 4 a$ , then it is easy to prove that the problem is equivalent to finding all possible values of $a$ such that
$\exists x \in C, ∣ x ∣ = 3 \land f (x) = 0$ $\exists x \in C, ∣ x ∣ = 3 \land f (x) = 0 ⟺ (\exists x \in R, ∣ x ∣ = 3 \land f (x) = 0) \lor (\exists x \in C ∖ R, ∣ x ∣ = 3 \land f (x) = 0)$ $\exists x \in R, ∣ x ∣ = 3 \land f (x) = 0 ⟺ \exists x \in {- 3, 3}, f (x) = 0 ⟺ a^{2} + 2 a + 9 = 0 \lor a^{2} - 10 a + 9 = 0 ⟺ a = 1 \lor a = 9$
Since $f (x) \in R [x]$ , it follows that
$f (x) = 0 ⟺ f (\overline{x}) = 0$
Therefore, applying Vieta’s Formulas yields
$\exists x \in C ∖ R, ∣ x ∣ = 3 \land f (x) = 0 ⟺ \exists x \in C ∖ R, ∣ x ∣ = 3 \land x + \overline{x} = 2 a \land x \overline{x} = a^{2} - 4 a ⟺ \exists x \in C ∖ R, ∣ x ∣ = 3 \land ℜ (x) = a \land ∣ x ∣^{2} = a^{2} - 4 a ⟺ a^{2} - 4 a = 9 \land \exists x \in C ∖ R, ∣ x ∣ = 3 \land ℜ (x) = a ⟺ a^{2} - 4 a - 9 = 0 \land - 3 \leq a \leq 3 ⟺ a = 2 - 13$
Therefore,
$\exists x \in C, ∣ x ∣ = 3 \land f (x) = 0 ⟺ a = 1 \lor a = 9 \lor a = 2 - 13$

Problem 16

Let $z = a + bi$ be the complex number with $∣ z ∣ = 5$ and $b > 0$ such that the distance between $(1 + 2 i) z^{3}$ and $z^{5}$ is maximized, and let $z^{4} = c + d i$ . Find $c + d$ .

Solution

$ar g ∣ z ∣ = 5 \land ℑ (z) > 0 max (1 + 2 i) z^{3} - z^{5} = ar g ∣ z ∣ = 5 \land ℑ (z) > 0 max ∣ z ∣^{3} 1 + 2 i - z^{2} = ar g ∣ z ∣ = 5 \land ℑ (z) > 0 max 1 + 2 i - z^{2} = {- 25 \frac{1 + 2 i}{∣1 + 2 i ∣}} = {- 55 (1 + 2 i)}$
Therefore,
$z = - 55 (1 + 2 i)$
Therefore,
$z^{4} = (- 55 (1 + 2 i))^{2} = - 375 + 500 i$
Therefore, $c + d = 125$ .

Problem 17

Evaluate

n = 0 \sum \infty \frac{cos ( n θ )}{2 ^{n}}

where $cos (θ) = \frac{1}{5}$ .

Solution

$n = 0 \sum \infty \frac{cos ( n θ )}{2 ^{n}} = n = 0 \sum \infty \frac{ℜ ( e ^{in θ} )}{2 ^{n}} = ℜ (n = 0 \sum \infty (\frac{e ^{i θ}}{2})^{n}) = 2ℜ (\frac{1}{2 - e ^{i θ}}) = 2ℜ (\frac{2 - cos ( θ ) + i sin ( θ )}{( 2 - cos ( θ ) ) ^{2} + sin ( θ ) ^{2}}) = \frac{4 - 2 cos ( θ )}{( 2 - cos ( θ ) ) ^{2} + 1 - cos ( θ ) ^{2}} = \frac{4 - 2 cos ( θ )}{( 2 - cos ( θ ) ) ^{2} + 1 - cos ( θ ) ^{2}} = \frac{6}{7}$

Problem 18

Let $ω \neq = 1$ be a $13$ th root of unity. Find the remainder when $\prod_{k = 0}^{12} (2 - 2 ω^{k} + ω^{2 k})$ is divided by $1000$ .

Solution

$k = 0 \prod 12 (2 - 2 ω^{k} + ω^{2 k}) = x^{13} - 1 = 0 \prod (x^{2} - 2 x + 2) = x^{13} - 1 = 0 \prod y^{2} - 2 y + 2 = 0 \prod (x - y) = y^{2} - 2 y + 2 = 0 \prod x^{13} - 1 = 0 \prod (y - x) = y^{2} - 2 y + 2 = 0 \prod (y^{13} - 1) = ((1 - i)^{13} - 1) ((1 + i)^{13} - 1) = 8321$
The remainder when $8321$ is divided by $1000$ is $321$ .

Problem 19

The polynomial $f (z) = a z^{2018} + b z^{2017} + c z^{2016}$ has real coefficients not exceeding $2019$ , and $f (\frac{1 + 3 i}{2}) = 2015 + 20193 i$ . Find the remainder when $f (1)$ is divided by $1000$ .

Solution

Since $(\frac{1 + 3 i}{2})^{6} = 1$ , it follows that
$f (\frac{1 + 3 i}{2}) = a (\frac{1 + 3 i}{2})^{2018} + b (\frac{1 + 3 i}{2})^{2017} + c (\frac{1 + 3 i}{2})^{2016} = a (\frac{- 1 + 3 i}{2}) + b (\frac{1 + 3 i}{2}) + c = - \frac{a}{2} + \frac{b}{2} + c + (\frac{a}{2} + \frac{b}{2}) 3 i$
Therefore,
$f (\frac{1 + 3 i}{2}) = 2015 + 20193 i ⟺ - \frac{a}{2} + \frac{b}{2} + c + (\frac{a}{2} + \frac{b}{2}) 3 i = 2015 + 20193 i ⟺ - \frac{a}{2} + \frac{b}{2} + c = 2015 \land \frac{a}{2} + \frac{b}{2} = 2019 ⟺ a = 2019 \land b = 2019 \land c = 2015$
Therefore,
$f (1) = a + b + c = 6053$
The remainder when $6053$ is divided by $1000$ is $53$ .

Problem 20

Find the largest possible real part of $(75 + 117 i) z + \frac{96 + 144 i}{z}$ where $z$ is a complex number with $∣ z ∣ = 4$ .

Solution

$∣ z ∣ = 4 max ℜ ((75 + 117 i) z + \frac{96 + 144 i}{z}) = θ max ℜ ((300 + 468 i) e^{i θ} + (24 + 36 i) e^{- i θ}) = θ max (324 cos (θ) - 432 sin (θ))$
Applying Cauchy-Schwarz Inequality yields
$324 cos (θ) - 432 sin (θ) \leq (32 4^{2} + 43 2^{2}) (cos (θ)^{2} + sin (θ)^{2}) = 540$
and when $θ = - arctan (\frac{4}{3})$ , $324 cos (θ) - 432 sin (θ) = 540$ . Therefore,
$θ max (324 cos (θ) - 432 sin (θ)) = 540$

Problem 21

Let $a, b, c, d$ be real numbers such that $b - d \geq 5$ and all zeros $x_{1}, x_{2}, x_{3}, x_{4}$ of the polynomial

P (x) = x^{4} + a x^{3} + b x^{2} + c x + d

are real. Find the smallest possible value of the product

(x_{1}^{2} + 1) (x_{2}^{2} + 1) (x_{3}^{2} + 1) (x_{4}^{2} + 1)

Solution

$(x_{1}^{2} + 1) (x_{2}^{2} + 1) (x_{3}^{2} + 1) (x_{4}^{2} + 1) = (i - x_{1}) (- i - x_{1}) (i - x_{2}) (- i - x_{2}) (i - x_{3}) (- i - x_{3}) (i - x_{4}) (- i - x_{4}) = P (i) P (- i) = (b - d - 1 + (a - c) i) (b - d + 1 - (a - c) i) = (b - d - 1)^{2} + (a - c)^{2}$
Therefore, the smallest possible value of the product is $16$ .

Problem 22

Let $N$ be the number of complex numbers $z$ with the properties that $∣ z ∣ = 1$ and $z^{6!} - z^{5!}$ is a real number. Find the remainder when $N$ is divided by $1000$ .

Solution

$N = ∣ {z : z \in C \land ∣ z ∣ = 1 \land z^{6!} - z^{5!} \in R} ∣ = ∣ {θ : θ \in [0, 2 π) \land e^{6! i θ} - e^{5! i θ} \in R} ∣$ $e^{6! i θ} - e^{5! i θ} \in R ⟺ sin (6! θ) - sin (5! θ) = 0 ⟺ 2 cos (420 θ) sin (300 θ) = 0 ⟺ cos (420 θ) = 0 \lor sin (300 θ) = 0 ⟺ (\exists n \in Z, θ = \frac{( 2 n + 1 ) π}{840}) \lor (\exists n \in Z, θ = \frac{nπ}{300})$
Therefore,
$N = {\frac{1 π}{840}, \frac{3 π}{840}, \dots, \frac{1679 π}{840}} \cup {\frac{0 π}{300}, \frac{1 π}{300}, \dots, \frac{599 π}{300}} = ∣ {1, 3, \dots, 1679} ∣ + ∣ {0, 1, \dots, 599} ∣ = 1440$
Therefore, the remainder when $N$ is divided by $1000$ is $440$ .

Problem 23

Evaluate the following sums:

(a)

(0 48) + (3 48) + \dots + (48 48)

Solution

$(0 48) + (3 48) + \dots + (48 48) = n = 0 \sum 48 (n 48) [3 ∣ n] = \frac{1}{3} n = 0 \sum 48 (n 48) k = 0 \sum 2 ω_{3}^{kn} = \frac{1}{3} k = 0 \sum 2 n = 0 \sum 48 (n 48) (ω_{3}^{k})^{n} = \frac{1}{3} k = 0 \sum 2 (1 + ω_{3}^{k})^{48} = \frac{2 ^{48} + ( \frac{1 + 3 i}{2} ) ^{48} + ( \frac{1 - 3 i}{2} ) ^{48}}{3} = \frac{2 ^{48} + 2}{3}$

(b)

(0 48) + (4 48) + \dots + (48 48)

Solution

$(0 48) + (4 48) + \dots + (48 48) = n = 0 \sum 48 (n 48) [4 ∣ n] = \frac{1}{4} n = 0 \sum 48 (n 48) k = 0 \sum 3 ω_{4}^{kn} = \frac{1}{4} k = 0 \sum 3 n = 0 \sum 48 (n 48) (ω_{4}^{k})^{n} = \frac{1}{4} k = 0 \sum 3 (1 + ω_{4}^{k})^{48} = \frac{2 ^{48} + ( 1 + i ) ^{48} + 0 ^{48} + ( 1 - i ) ^{48}}{4} = \frac{2 ^{48} + 2 ^{25}}{4}$

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