PRISMS Senior Varsity Training 20250922

news/2025/10/20 2:54:17/文章来源:https://www.cnblogs.com/rdfz-xd/p/-/8

Problem 1

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

\[\log(kx)=2\log(x+2) \]

has exactly one real solution.

Solution 1

\[\begin{align} \log(kx)=2\log(x+2)&\Longleftrightarrow x>-2\land\log(kx)=\log((x+2)^2)\\ &\Longleftrightarrow x>-2\land kx>0\land kx=(x+2)^2\\ &\Longleftrightarrow x>-2\land kx>0\land x^2+(4-k)x+4=0\\ \end{align} \]

Let \(f(x)=x^2+(4-k)x+4=(x-x_1)(x-x_2)\), where \(x_1=\frac{k-4-\sqrt{k(k-8)}}{2},x_2=\frac{k-4+\sqrt{k(k-8)}}{2}\), then

\[\begin{align} \log(kx)=2\log(x+2)&\Longleftrightarrow x>-2\land kx>0\land f(x)=0\\ &\Longleftrightarrow x>-2\land kx>0\land(x=x_1\lor x=x_2)\\ &\Longrightarrow k<0\lor k\ge8 \end{align} \]

Applying Vieta's Formulas yields

\[x_1+x_2=k-4\land x_1x_2=4 \]

If \(k<0\),

\[\begin{align} x_1+x_2=k-4\land x_1x_2=4&\Longrightarrow x_1+x_2<0\land x_1x_2>0\\ &\Longleftrightarrow x_1<0\land x_2<0\\ \end{align} \]

Since \(f''(x)=2>0\land f(-2)=2k<0\), it follows that

\[x_1<-2\land x_2>-2 \]

Therefore,

\[\log(kx)=2\log(x+2)\Longleftrightarrow x=x_2 \]

Therefore, if \(k<0\), \(\log(kx)=2\log(x+2)\) always has exactly one real solution.

If \(k\ge8\),

\[\begin{align} x_1+x_2=k-4\land x_1x_2=4&\Longrightarrow x_1+x_2>0\land x_1x_2>0\\ &\Longleftrightarrow x_1>0\land x_2>0\\ \end{align} \]

Therefore,

\[\log(kx)=2\log(x+2)\Longleftrightarrow x=x_1\lor x=x_2 \]

Therefore, if \(k\ge8\), \(\log(kx)=2\log(x+2)\) has exactly one real solution iff

\[\begin{align} x_1=x_2&\Longleftrightarrow k(k-8)=0\\ &\Longleftrightarrow k=8 \end{align} \]

Therefore, \(\log(kx)=2\log(x+2)\) 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 \(\log_x(y^x)=\log_y(x^{4y})=10\). What is the value of \(xy\)?

Solution 2

\[\begin{align} \log_x(y^x)\log_y(x^{4y})&=x\log_x(y)\cdot4y\log_y(x)\\ &=4xy \end{align} \]

Therefore,

\[4xy=100 \]

\[4xy=100\Longleftrightarrow xy=25 \]

Problem 3

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

\[\log_2\left(\frac{x}{yz}\right)=\frac{1}{2},\log_2\left(\frac{y}{xz}\right)=\frac{1}{3},\log_2\left(\frac{z}{xy}\right)=\frac{1}{4} \]

Then the value of \(\left|\log_2(x^4y^3z^2)\right|\) is \(\frac{m}{n}\) where \(m\) and \(n\) are relatively prime positive integers. Find \(m+n\).

Solution 3

\[\begin{align} \log_2\left(\frac{x}{yz}\right)=\frac{1}{2}\land\log_2\left(\frac{y}{xz}\right)=\frac{1}{3}\land\log_2\left(\frac{z}{xy}\right)=\frac{1}{4}&\Longleftrightarrow\log_2(x)-\log_2(y)-\log_2(z)=\frac{1}{2}\land\log_2(y)-\log_2(x)-\log_2(z)=\frac{1}{3}\land\log_2(z)-\log_2(x)-\log_2(y)=\frac{1}{4}\\ &\Longleftrightarrow\log_2(x)=-\frac{7}{24}\land\log_2(y)=-\frac{3}{8}\land\log_2(z)=-\frac{5}{12} \end{align} \]

Therefore,

\[\begin{align} \left|\log_2(x^4y^3z^2)\right|&=\left|4\log_2(x)+3\log_2(y)+2\log(z)\right|\\ &=\frac{25}{8} \end{align} \]

Therefore, \(m+n=33\).

Problem 4

Let \(a>1\) and \(x>1\) satisfy

\[\log_a(\log_a(\log_a(2))+\log_a(2)-128)=128\land\log_a(\log_a(x))=256 \]

Find the reminder when \(x\) is divided by \(1000\).

Solution 4

\[\begin{align} \log_a(\log_a(\log_a(2))+\log_a(24)-128)=128&\Longleftrightarrow\log_a(24\log_a(2))=128+a^{128}\\ &\Longleftrightarrow\log_a(2^{24})=a^{128}a^{a^{128}}\\ &\Longleftrightarrow2^{24}=\left(a^{a^{128}}\right)^{a^{a^{128}}}\\ &\Longleftrightarrow a^{a^{128}}=2^3\\ &\Longleftrightarrow a=2^\frac{3}{64}\\ \end{align} \]

Therefore,

\[\begin{align} \log_a(\log_a(x))=256&\Longleftrightarrow x=a^{a^{256}}\\ &\Longleftrightarrow x=2^{192}\\ &\Longrightarrow x\equiv896\pmod{1000} \end{align} \]

Problem 5

Let \(x\), \(y\) and \(z\) be real numbers satisfying the system

\[\log_2(xyz-3+\log_5(x))=5\\ \log_3(xyz-3+\log_5(y))=4\\ \log_4(xyz-3+\log_5(z))=4 \]

Find the value of \(\left|\log_5(x)\right|+\left|\log_5(y)\right|+\left|\log_5(z)\right|\).

Solution 5

\[\begin{align} \log_2(xyz-3+\log_5(x))=5\land\log_3(xyz-3+\log_5(y))=4\land\log_4(xyz-3+\log_5(z))=4&\Longleftrightarrow xyz+\log_5(x)=35\land xyz+\log_5(y)=84\land xyz+\log_5(z)=259\\ &\Longrightarrow3xyz+\log_5(xyz)=378\\ &\Longleftrightarrow xyz=125 \end{align} \]

Therefore,

\[xyz+\log_5(x)=35\Longleftrightarrow\log_5(x)=-90 \]

\[xyz+\log_5(y)=84\Longleftrightarrow\log_5(y)=-41 \]

\[xyz+\log_5(z)=259\Longleftrightarrow\log_5(z)=134 \]

Therefore,

\[\left|\log_5(x)\right|+\left|\log_5(y)\right|+\left|\log_5(z)\right|=265 \]

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 6

Let \(P(x)=2x^2+b_Px+c_P\), then

\[\begin{align} P(16)=54\land P(20)=53&\Longleftrightarrow512+16b_P+c_P=54\land800+20b_P+c_P=53\\ &\Longleftrightarrow b_P=-\frac{289}{4}\land c_P=698 \end{align} \]

Therefore,

\[P(x)=2x^2-\frac{289}{4}x+698 \]

Let \(Q(x)=-2x^2+b_Qx+c_Q\), then

\[\begin{align} Q(16)=54\land Q(20)=53&\Longleftrightarrow-512+16b_Q+c_Q=54\land-800+20b_Q+c_Q=53\\ &\Longleftrightarrow b_Q=\frac{287}{4}\land c_Q=-582 \end{align} \]

Therefore,

\[Q(x)=-2x^2+\frac{287}{4}x-582 \]

Therefore,

\[\begin{align} P(0)+Q(0)&=698-582\\ &=116 \end{align} \]

Problem 7

For certain real numbers \(a\), \(b\), and \(c\), the polynomial

\[g(x)=x^3+ax^2+x+10 \]

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

\[f(x)=x^4+x^3+bx^2+100x+c \]

Solution 7

Let \(x_0\) be the additional root of \(f(x)\), then

\[\begin{align} f(x)&=(x-x_0)g(x)\\ &=(x-x_0)(x^3+ax^2+x+10)\\ &=x^4+(a-x_0)x^3+(1-ax_0)x^2+(10-x_0)x-10x_0 \end{align} \]

Therefore,

\[a-x_0=1\land1-ax_0=b\land10-x_0=100\land-10x_0=c \]

\[a-x_0=1\land1-ax_0=b\land10-x_0=100\land-10x_0=c\Longleftrightarrow x_0=-90\land a=-89\land b=-8009\land c=900 \]

Therefore,

\[f(x)=x^4+x^3-8009x^2+100x+900 \]

Therefore, \(f(1)=-7007\).

Problem 8

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

Solution 8

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

\[\exist r,s\in\Z\setminus\{0\},P(r+si)=0\land\exist a,b\in\Z\setminus\{0\},P(x)=x^3-ax^2+bx-65 \]

Since \(P(x)\in\R[x]\), it follows that

\[P(r+si)=0\Longleftrightarrow P(r-si)=0 \]

Therefore,

\[\begin{align} \exist r,s\in\Z\setminus\{0\},P(r+si)=0&\Longleftrightarrow\exist r,s\in\Z\setminus\{0\},\exist t\in\R,P(x)=(x-r+si)(x-r-si)(x-t)\\ &\Longleftrightarrow\exist r,s\in\Z\setminus\{0\},\exist t\in\R,P(x)=x^3-(2r+t)x^2+(r^2+s^2+2rt)x-(r^2+s^2)t \end{align} \]

Therefore,

\[\begin{align} \exist r,s\in\Z\setminus\{0\},P(r+si)=0\land\exist a,b\in\Z\setminus\{0\},P(x)=x^3-ax^2+bx-65&\Longleftrightarrow\exist r,s\in\Z\setminus\{0\},\exist t\in\R,P(x)=x^3-(2r+t)x^2+(r^2+s^2+2rt)x-(r^2+s^2)t\land2r+t\in\Z\setminus\{0\}\land r^2+s^2+2rt\in\Z\setminus\{0\}\land(r^2+s^2)t=65\\ &\Longleftrightarrow\exist r,s\in\Z\setminus\{0\},\exist t\in\Z_+,P(x)=x^3-(2r+t)x^2+(r^2+s^2+2rt)x-(r^2+s^2)t\land(r^2+s^2)t=65\\ &\Longleftrightarrow\exist\langle r,s,t\rangle\in\{\langle1,2,13\rangle,\langle2,1,13\rangle,\langle-1,-2,13\rangle,\langle-2,-1,13\rangle,\langle2,3,5\rangle,\langle3,2,5\rangle,\langle-2,-3,5\rangle,\langle-3,-2,5\rangle,\langle1,8,1\rangle,\langle8,1,1\rangle,\langle-1,-8,1\rangle,\langle-8,-1,1\rangle\,\langle4,7,1\rangle,\langle7,4,1\rangle,\langle-4,-7,1\rangle,\langle-7,-4,1\rangle\},P(x)=(x-r+si)(x-r-si)(x-t) \end{align} \]

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 9

\[\begin{align} f(x^2)=f(x)^2&\Longrightarrow\forall x\in\mathbb{C},f(x^2)=0\leftrightarrow f(x)^2=0\\ &\Longleftrightarrow\forall x\in\mathbb{C},f(x^2)=0\leftrightarrow f(x)=0 \end{align} \]

Therefore, if \(\exist x\in\C\setminus\{-1,0,1\},f(x)=0\), infinitely many distinct numbers

\[x,x^2,x^4,x^8,\dots \]

are all zeros of \(f(x)\). If \(\exist x\in\{0,1\},f(x)=0\), infinitely many distinct numbers

\[e^{i\pi},e^\frac{i\pi}{2},e^\frac{i\pi}{4},e^\frac{i\pi}{8},\dots \]

are all zeros of \(f(x)\). Therefore,

\[\begin{align} f(x^2)=f(x)^2&\Longrightarrow f(x)=0\lor\forall x\in\mathbb{C}\setminus\{0\},f(x)\ne0\\ &\Longleftrightarrow\exist a\in\R,\exist n\in\N,f(x)=ax^n \end{align} \]

Therefore,

\[\begin{align} f(x^2)=f(x)^2&\Longleftrightarrow\exist a\in\R,\exist n\in\N,f(x)=ax^n\land ax^{2n}=a^2x^{2n}\\ &\Longleftrightarrow\exist a\in\R,\exist n\in\N,f(x)=ax^n\land(a=0\lor a=1)\\ &\Longleftrightarrow f(x)=0\lor\exist n\in\N,f(x)=x^n \end{align} \]

Problem 10

There exist two triples of real numbers \(\langle a,b,c\rangle\) such that \(a-\frac{1}{b}\), \(b-\frac{1}{c}\), and \(c-\frac{1}{a}\) are the roots to the cubic equation \(x^3-5x^2-15x+3\) listed in increasing order. Denote those \(\langle a_1,b_1,c_1\rangle\) and \(\langle a_2,b_2,c_2\rangle\). 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 10

Applying Vieta's Lemma yields

\[\left(a-\frac{1}{b}\right)\left(b-\frac{1}{c}\right)\left(c-\frac{1}{a}\right)=-3 \]

\[\left(a-\frac{1}{b}\right)\left(b-\frac{1}{c}\right)\left(c-\frac{1}{a}\right)=-3\Longleftrightarrow abc-a-b-c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=-3 \]

Applying Vieta's Lemma yields

\[a-\frac{1}{b}+b-\frac{1}{c}+c-\frac{1}{a}=5 \]

Therefore,

\[abc-\frac{1}{abc}=2 \]

\[\begin{align} abc-\frac{1}{abc}=2&\Longleftrightarrow(abc)^2-2abc-1=0\\ &\Longleftrightarrow abc=1-\sqrt2\lor abc=1+\sqrt2 \end{align} \]

Applying Vieta's Lemma yields

\[f(0)=-a_1b_1c_1,g(0)=-a_2b_2c_2 \]

It is easy to prove that \(f(0)\ne g(0)\), therefore

\[\begin{align} f(0)^3+g(0)^3&=\left(-\left(1-\sqrt2\right)\right)^3+\left(-\left(1+\sqrt2\right)\right)^3\\ &=-14 \end{align} \]

Problem 11

Find all possible values of \(a\) such that the roots of polynomial \(x^3-6x^2+ax+a\) (denote by \(x_1,x_2,x_3\)) satisfy that \((x_1-3)^3+(x_2-3)^3+(x_3-3)^3=0\).

Solution 11

Applying Newton's Identities yields

\[x_1+x_2+x_3=6 \]

\[\begin{align} x_1^2+x_2^2+x_3^2&=6(x_1+x_2+x_3)-2a\\ &=36-2a \end{align} \]

\[\begin{align} x_1^3+x_2^3+x_3^3&=6(x_1^2+x_2^2+x_3^2)-a(x_1+x_2+x_3)-3a\\ &=216-21a \end{align} \]

Therefore,

\[\begin{align} (x_1-3)^3+(x_2-3)^3+(x_3-3)^3&=x_1^3+x_2^3+x_3^3-9(x_1^2+x_2^2+x_3^2)+27(x_1+x_2+x_3)-81\\ &=-3a-27 \end{align} \]

Therefore,

\[\begin{align} (x_1-3)^3+(x_2-3)^3+(x_3-3)^3=0&\Longleftrightarrow-3a-27=0\\ &\Longleftrightarrow a=-9 \end{align} \]

Problem 12

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

Solution 12

Let

\[\begin{align} g(x)&=x^nf(x^{-1})\\ &=2x^4+3x^3+3x+1 \end{align} \]

then, since

\[\begin{align} g(x)&=x^nf(x^{-1})\\ &=(rx-1)(sx-1)(tx-1)(ux-1) \end{align} \]

\(g(x)\) is a polynomial with roots \(\frac{1}{r}\), \(\frac{1}{s}\), \(\frac{1}{t}\) and \(\frac{1}{u}\).

Applying Newton's Identities yields

\[\frac{1}{r}+\frac{1}{s}+\frac{1}{t}+\frac{1}{u}=-\frac{3}{2} \]

\[\begin{align} \frac{1}{r^2}+\frac{1}{s^2}+\frac{1}{t^2}+\frac{1}{u^2}&=-\frac{3}{2}\left(\frac{1}{r}+\frac{1}{s}+\frac{1}{t}+\frac{1}{u}\right)\\ &=\frac{9}{4} \end{align} \]

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 13

It is easy to prove the problem is equivalent to finding all polynomials \(f(x)\) such that

\[\exist a,b,c,d\in\Z,f(x)=ax^3+bx^2+cx+d\land f(0)=1\land f(1)=1\land f(2)=5\land f(3)=11 \]

\[\begin{align} \exist a,b,c,d\in\Z,f(x)=ax^3+bx^2+cx+d\land f(0)=1\land f(1)=1\land f(2)=5\land f(3)=11&\Longleftrightarrow\exist a,b,c,d\in\Z,f(x)=ax^3+bx^2+cx+d\land d=1\land a+b+c+d=1\land 8a+4b+2c+d=5\land 27a+9b+3c+d=11\\ &\Longleftrightarrow\exist a,b,c,d\in\Z,f(x)=ax^3+bx^2+cx+d\land a=-\frac{1}{3}\land b=3\land c=-\frac{8}{3}\land d=1\\ &\Longleftrightarrow f(x)=-\frac{1}{3}x^3+3x^2-\frac{8}{3}x+1 \end{align} \]

Problem 14

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

Solution 14

The roots of \(z^{12}=64\) are

\[z_k=\sqrt2\left(\cos\left(\frac{k\pi}{6}\right)+i\sin\left(\frac{k\pi}{6}\right)\right),k=0,1,\dots,11 \]

Since

\[\cos\left(\frac{k\pi}{6}\right)>0\Longleftrightarrow k\in\{0,1,2,10,11\} \]

It follows that the roots with a positive real part are \(z_0,z_1,z_2,z_{10},z_{11}\).

\[\begin{align} \sum_{k\in\{0,1,2,10,11\}}z_k&=\sqrt2\sum_{k\in\{0,1,2,10,11\}}\cos\left(\frac{k\pi}{6}\right)+\sqrt2i\sum_{k\in\{0,1,2,10,11\}}\sin\left(\frac{k\pi}{6}\right)\\ &=\sqrt2\left(\cos(0)+2\cos\left(\frac{\pi}{6}\right)+2\cos\left(\frac{\pi}{3}\right)\right)\\ &=2\sqrt2+\sqrt6 \end{align} \]

Problem 15

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

Solution 15

Let \(f(x)=x^2-2ax+a^2-4a\), then it is easy to prove that the problem is equivalent to finding all possible values of \(a\) such that

\[\exist x\in\mathbb C,|x|=3\land f(x)=0 \]

\[\exist x\in\mathbb C,|x|=3\land f(x)=0\Longleftrightarrow(\exist x\in\R,|x|=3\land f(x)=0)\lor(\exist x\in\mathbb C\setminus\R,|x|=3\land f(x)=0) \]

\[\begin{align} \exist x\in\R,|x|=3\land f(x)=0&\Longleftrightarrow\exist x\in\{-3,3\},f(x)=0\\ &\Longleftrightarrow a^2+2a+9=0\lor a^2-10a+9=0\\ &\Longleftrightarrow a=1\lor a=9\\ \end{align} \]

Since \(f(x)\in\R[x]\), it follows that

\[f(x)=0\Longleftrightarrow f(\overline x)=0 \]

Therefore, applying Vieta's Formulas yields

\[\begin{align} \exist x\in\mathbb C\setminus\R,|x|=3\land f(x)=0&\Longleftrightarrow \exist x\in\mathbb C\setminus\R,|x|=3\land x+\overline{x}=2a\land x\overline{x}=a^2-4a\\ &\Longleftrightarrow \exist x\in\mathbb C\setminus\R,|x|=3\land\Re(x)=a\land|x|^2=a^2-4a\\ &\Longleftrightarrow a^2-4a=9\land\exist x\in\mathbb C\setminus\R,|x|=3\land\Re(x)=a\\ &\Longleftrightarrow a^2-4a-9=0\land-3\le a\le3\\ &\Longleftrightarrow a=2-\sqrt{13} \end{align} \]

Therefore,

\[\exist x\in\mathbb C,|x|=3\land f(x)=0\Longleftrightarrow a=1\lor a=9\lor a=2-\sqrt{13} \]

Problem 16

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

Solution 16

\[\begin{align} \operatorname*{argmax}_{|z|=5\land\Im(z)>0}\left|(1+2i)z^3-z^5\right|&=\operatorname*{argmax}_{|z|=5\land\Im(z)>0}|z|^3\left|1+2i-z^2\right|\\ &=\operatorname*{argmax}_{|z|=5\land\Im(z)>0}\left|1+2i-z^2\right|\\ &=\left\{\sqrt{-25\frac{1+2i}{|1+2i|}}\right\}\\ &=\left\{\sqrt{-5\sqrt5(1+2i)}\right\} \end{align} \]

Therefore,

\[z=\sqrt{-5\sqrt5(1+2i)} \]

Therefore,

\[\begin{align} z^4&=\left(-5\sqrt5(1+2i)\right)^2\\ &=-375+500i \end{align} \]

Therefore, \(c+d=125\).

Problem 17

Evaluate

\[\sum_{n=0}^\infty\frac{\cos(n\theta)}{2^n} \]

where \(\cos(\theta)=\frac{1}{5}\).

Solution 17

\[\begin{align} \sum_{n=0}^\infty\frac{\cos(n\theta)}{2^n}&=\sum_{n=0}^\infty\frac{\Re(e^{in\theta})}{2^n}\\ &=\Re\left(\sum_{n=0}^\infty\left(\frac{e^{i\theta}}{2}\right)^n\right)\\ &=2\Re\left(\frac{1}{2-e^{i\theta}}\right)\\ &=2\Re\left(\frac{2-\cos(\theta)+i\sin(\theta)}{(2-\cos(\theta))^2+\sin(\theta)^2}\right)\\ &=\frac{4-2\cos(\theta)}{(2-\cos(\theta))^2+1-\cos(\theta)^2}\\ &=\frac{4-2\cos(\theta)}{(2-\cos(\theta))^2+1-\cos(\theta)^2}\\ &=\frac{6}{7} \end{align} \]

Problem 18

Let \(\omega\ne1\) be a \(13\)th root of unity. Find the remainder when \(\prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})\) is divided by \(1000\).

Solution 18

\[\begin{align} \prod_{k=0}^{12}(2-2\omega^k+\omega^{2k})&=\prod_{x^{13}-1=0}(x^2-2x+2)\\ &=\prod_{x^{13}-1=0}\prod_{y^2-2y+2=0}(x-y)\\ &=\prod_{y^2-2y+2=0}\prod_{x^{13}-1=0}(y-x)\\ &=\prod_{y^2-2y+2=0}(y^{13}-1)\\ &=((1-i)^{13}-1)((1+i)^{13}-1)\\ &=8321 \end{align} \]

The remainder when \(8321\) is divided by \(1000\) is \(321\).

Problem 19

The polynomial \(f(z)=az^{2018}+bz^{2017}+cz^{2016}\) has real coefficients not exceeding \(2019\), and \(f\left(\frac{1+\sqrt3i}{2}\right)=2015+2019\sqrt3i\). Find the remainder when \(f(1)\) is divided by \(1000\).

Solution 19

Since \(\left(\frac{1+\sqrt3i}{2}\right)^6=1\), it follows that

\[\begin{align} f\left(\frac{1+\sqrt3i}{2}\right)&=a\left(\frac{1+\sqrt3i}{2}\right)^{2018}+b\left(\frac{1+\sqrt3i}{2}\right)^{2017}+c\left(\frac{1+\sqrt3i}{2}\right)^{2016}\\ &=a\left(\frac{-1+\sqrt3i}{2}\right)+b\left(\frac{1+\sqrt3i}{2}\right)+c\\ &=-\frac{a}{2}+\frac{b}{2}+c+\left(\frac{a}{2}+\frac{b}{2}\right)\sqrt3i \end{align} \]

Therefore,

\[\begin{align} f\left(\frac{1+\sqrt3i}{2}\right)2015+2019\sqrt3i&\Longleftrightarrow-\frac{a}{2}+\frac{b}{2}+c+\left(\frac{a}{2}+\frac{b}{2}\right)\sqrt3i=2015+2019\sqrt3i\\ &\Longleftrightarrow-\frac{a}{2}+\frac{b}{2}+c=2015\land\frac{a}{2}+\frac{b}{2}=2019\\ &\Longleftrightarrow a=2019\land b=2019\land c=2015\\ \end{align} \]

Therefore,

\[\begin{align} f(1)&=a+b+c\\ &=6053 \end{align} \]

The remainder when \(6053\) is divided by \(1000\) is \(53\).

Problem 20

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

Solution 20

\[\begin{align} \max_{|z|=4}\Re\left((75+117i)z+\frac{96+144i}{z}\right)&=\max_\theta\Re((300+468i)e^{i\theta}+(24+36i)e^{-i\theta})\\ &=\max_\theta(324\cos(\theta)-432\sin(\theta)) \end{align} \]

Applying Cauchy-Schwarz Inequality yields

\[\begin{align} 324\cos(\theta)-432\sin(\theta)&\le\sqrt{(324^2+432^2)(\cos(\theta)^2+\sin(\theta)^2)}\\ &=540 \end{align} \]

and when \(\theta=-\arctan\left(\frac{4}{3}\right)\), \(324\cos(\theta)-432\sin(\theta)=540\). Therefore,

\[\max_\theta(324\cos(\theta)-432\sin(\theta))=540 \]

Problem 21

Let \(a,b,c,d\) be real numbers such that \(b-d\ge5\) and all zeros \(x_1,x_2,x_3,x_4\) of the polynomial

\[P(x)=x^4+ax^3+bx^2+cx+d \]

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 21

\[\begin{align} (x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)&=(i-x_1)(-i-x_1)(i-x_2)(-i-x_2)(i-x_3)(-i-x_3)(i-x_4)(-i-x_4)\\ &=P(i)P(-i)\\ &=(b-d-1+(a-c)i)(b-d+1-(a-c)i)\\ &=(b-d-1)^2+(a-c)^2 \end{align} \]

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 22

\[\begin{align} N&=|\{z:z\in\mathbb{C}\land|z|=1\land z^{6!}-z^{5!}\in\R\}|\\ &=|\{\theta:\theta\in[0,2\pi)\land e^{6!i\theta}-e^{5!i\theta}\in\R\}| \end{align} \]

\[\begin{align} e^{6!i\theta}-e^{5!i\theta}\in\R&\Longleftrightarrow\sin(6!\theta)-\sin(5!\theta)=0\\ &\Longleftrightarrow2\cos(420\theta)\sin(300\theta)=0\\ &\Longleftrightarrow\cos(420\theta)=0\lor\sin(300\theta)=0\\ &\Longleftrightarrow\left(\exist n\in\Z,\theta=\frac{(2n+1)\pi}{840}\right)\lor\left(\exist n\in\Z,\theta=\frac{n\pi}{300}\right) \end{align} \]

Therefore,

\[\begin{align} N&=\left|\left\{\frac{1\pi}{840},\frac{3\pi}{840},\dots,\frac{1679\pi}{840}\right\}\cup\left\{\frac{0\pi}{300},\frac{1\pi}{300},\dots,\frac{599\pi}{300}\right\}\right|\\ &=|\{1,3,\dots,1679\}|+|\{0,1,\dots,599\}|\\ &=1440 \end{align} \]

Therefore, the remainder when \(N\) is divided by \(1000\) is \(440\).

Problem 23

Evaluate the following sums:

(a)

\[\binom{48}{0}+\binom{48}{3}+\dots+\binom{48}{48} \]

(b)

\[\binom{48}{0}+\binom{48}{4}+\dots+\binom{48}{48} \]

Solution 23

(a)

\[\begin{align} \binom{48}{0}+\binom{48}{3}+\dots+\binom{48}{48}&=\sum_{n=0}^{48}\binom{48}{n}[3\mid n]\\ &=\frac{1}{3}\sum_{n=0}^{48}\binom{48}{n}\sum_{k=0}^2\omega_3^{kn}\\ &=\frac{1}{3}\sum_{k=0}^2\sum_{n=0}^{48}\binom{48}{n}(\omega_3^k)^n\\ &=\frac{1}{3}\sum_{k=0}^2(1+\omega_3^k)^{48}\\ &=\frac{2^{48}+\left(\frac{1+\sqrt3i}{2}\right)^{48}+\left(\frac{1-\sqrt3i}{2}\right)^{48}}{3}\\ &=\frac{2^{48}+2}{3}\\ \end{align} \]

(b)

\[\begin{align} \binom{48}{0}+\binom{48}{4}+\dots+\binom{48}{48}&=\sum_{n=0}^{48}\binom{48}{n}[4\mid n]\\ &=\frac{1}{4}\sum_{n=0}^{48}\binom{48}{n}\sum_{k=0}^3\omega_4^{kn}\\ &=\frac{1}{4}\sum_{k=0}^3\sum_{n=0}^{48}\binom{48}{n}(\omega_4^k)^n\\ &=\frac{1}{4}\sum_{k=0}^3(1+\omega_4^k)^{48}\\ &=\frac{2^{48}+(1+i)^{48}+0^{48}+(1-i)^{48}}{4}\\ &=\frac{2^{48}+2^{25}}{4}\\ \end{align} \]

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