Square both sides.

Rearrange and use Identity 10 (Difference of Squares).

n^2-m^2=(n-m)(n+m)=49

n-m and n+m are factors of 49.

(B) 2

Use 30. Vieta’s Formula.

\text{Volume} = (r_1 + 2)(r_2 + 2)(r_3 + 2)

By using 30. Vieta’s Formula:
\begin{aligned} r_1 + r_2 + r_3 &= \frac{39}{10}, \\ r_1r_2 + r_1r_3 + r_2r_3 &= \frac{29}{10}, \\ r_1r_2r_3 &= \frac{6}{10}. \end{aligned}

Expand:
(r_1 + 2)(r_2 + 2)(r_3 + 2) = r_1r_2r_3 + 2(r_1r_2 + r_2r_3 + r_3r_1) + 4(r_1 + r_2 + r_3) + 8. Substitute from Vieta’s results:
\frac{6}{10} + 2\left(\frac{29}{10}\right) + 4\left(\frac{39}{10}\right) + 8 = 30.

(D) 30

Use 31. Signs and Roots of Quadratics.

\[ \Delta \le 0 \Rightarrow b^2-4c \le 0 \Rightarrow b^2 \le 4c, \quad c^2-4b \le 0 \Rightarrow c^2 \le 4b. \]

From above, \[ b^2 \le 4c \text{ and } c^2 \le 4b \Rightarrow b^4 \le 16c^2 \le 64b \Rightarrow b^3 \le 64 \Rightarrow b \le 4. \] Similarly \(c \le 4\).

Case \(b=3\): \(9 \le 4c \Rightarrow c=3,\) and \(c^2 \le 12\) holds.

Case \(b=4\): \(16 \le 4c \Rightarrow c=4,\) and \(c^2 \le 16\) holds. Checking all valid \((b,c)\) up to 4 gives 6 pairs.

Final Answer: (B) 6

Use 10. Difference of squares.

Multiply by \((3-2)\).

\[ (3-2)(2+3)(2^2+3^2)(2^4+3^4)\cdots(2^{64}+3^{64}) =(3^2-2^2)(2^2+3^2)(2^4+3^4)\cdots \] \[ =(3^4-2^4)(2^4+3^4)\cdots =(3^8-2^8)(2^8+3^8)\cdots \Rightarrow \cdots = 3^{128}-2^{128}. \] So the original product is \((3^{128}-2^{128})/(3-2)=3^{128}-2^{128}.\)

(C) \(3^{128}-2^{128}\)

Use 30. Vieta’s Formula.

Sum of roots: \(r_1+\cdots+r_6=10.\) Product of roots: \(r_1r_2r_3r_4r_5r_6=16.\)

The only way with positive integers: roots \(2,2,2,2,1,1\).

By Vieta, the coefficient of \(z^3\) is \(-\!\sum r_ir_jr_k = -B\).

Count distinct triple products: \[ \binom{4}{3}(2\cdot2\cdot2) +\binom{4}{2}\binom{2}{1}(2\cdot2\cdot1) +\binom{4}{1}\binom{2}{2}(2\cdot1\cdot1) =32+48+8=88. \] So \(B=-88\).

(A) \(-88\)

Use the identity for the sum of odd powers (when the exponent is odd): \[ u^{2n+1} + 1 = (u+1)\big(u^{2n} – u^{2n-1} + u^{2n-2} – \cdots – u + 1\big). \] Here \(u = 2^{17}\) and \(2n+1 = 17\cdot17 = 289\) is odd.

Factor the numerator: \[ 2^{289} + 1 = \big(2^{17}\big)^{17} + 1 = \big(2^{17} + 1\big) \Big( (2^{17})^{16} – (2^{17})^{15} + (2^{17})^{14} – \cdots + (2^{17})^{2} – (2^{17})^{1} + 1 \Big). \] Therefore, \[ \frac{2^{289}+1}{2^{17}+1} = (2^{17})^{16} – (2^{17})^{15} + (2^{17})^{14} – \cdots + (2^{17})^{2} – (2^{17})^{1} + 1. \] This is an alternating sum of 17 terms, starting with positive \((2^{17})^{16}\) and ending with \(+1\).

Key fact we’ll reuse: \[ 2^n + 2^n = 2^{n+1}. \] This lets us rewrite differences like \((2^{17})^m – (2^{17})^{m-1}\) as a sum of distinct powers of \(2\) with consecutive exponents.

Look at one “block”: \[ (2^{17})^{2} – (2^{17})^{1} = 2^{34} – 2^{17}. \] Now expand the difference of powers in base 2: \[ 2^{34} – 2^{17} = 2^{34} + 2^{33} – 2^{33} – 2^{17} = 2^{34} + 2^{33} + \cdots + 2^{18} + 2^{17} – 2^{17}. \] More cleanly (using geometric series structure): \[ 2^{34} – 2^{17} = 2^{17}\big(2^{17}-1\big) = 2^{33} + 2^{32} + 2^{31} + \cdots + 2^{18} + 2^{17} – 2^{17} = 2^{33} + 2^{32} + \cdots + 2^{18}. \] This is a sum of 16 consecutive powers: \(2^{18}, 2^{19}, \ldots, 2^{33}.\) Including also that final lone \(+2^{17}\) from elsewhere will give us 17 distinct powers in total for that pair. In words: each adjacent pair \((2^{17})^{m} – (2^{17})^{m-1}\) “unpacks” into 17 distinct powers of 2 with consecutive exponents.

Similarly, for higher exponents: \[ (2^{17})^{4} – (2^{17})^{3} = 2^{68} – 2^{51} = 2^{67} + 2^{66} + \cdots + 2^{52} + 2^{51}. \] Again we get 17 distinct consecutive powers of 2.

The whole expression \[ (2^{17})^{16} – (2^{17})^{15} + (2^{17})^{14} – \cdots + (2^{17})^{2} – (2^{17})^{1} + 1 \] can be grouped into 8 “difference blocks” of the form \((2^{17})^{m} – (2^{17})^{m-1}\), each block expanding into 17 distinct powers of \(2\) with unique exponents, plus the final \(+1 = 2^0\). Because the exponents in different blocks don’t overlap, the binary expansion of the entire sum is just a bunch of distinct powers of 2, no carries needed. So we get: – one term from the final \(+1\), – plus \(8\) blocks, – with \(17\) distinct powers of \(2\) contributed per block.

Therefore \[ k = 1 + 17 \times 8 = 1 + 136 = 137. \]

(C) 137

Calculate x^2 + y^2.

By using 6. Square of a Sum: x^2 + y^2 = (x + y)^2 - 2xy = 4^2 - 2(-2) = 20

Calculate x^3 + y^3.

By using 12. Sum of cubes: x^3 + y^3 = (x + y)(x^2 - xy + y^2) = 4(20 + 2) = 88

Calculate x^5 + y^5.

x^5 + y^5 = (x^2 + y^2)(x^3 + y^3) - (x^2y^3 + x^3y^2) But x^2y^3 + x^3y^2 = x^2y^2(x + y). Since xy = -2 \Rightarrow x^2y^2 = 4, and x + y = 4, we have x^2y^2(x + y) = 4 \cdot 4 = 16. x^5 + y^5 = (20)(88) - 16 = 1760 - 16 = 1744

We want x + \dfrac{x^3}{y^2} + \dfrac{y^3}{x^2} + y = x + y + \dfrac{x^5 + y^5}{x^2y^2} x + y = 4, x^5 + y^5 = 1744, x^2y^2 = 4. \dfrac{x^5 + y^5}{x^2y^2} = \dfrac{1744}{4} = 436 Therefore, x + \dfrac{x^3}{y^2} + \dfrac{y^3}{x^2} + y = 4 + 436 = 440

(D) 440

Use 1. Sum of the first n integers.

Let S = 1 + 2 + 3 - 4 + 5 + 6 + 7 - 8 + \cdots + 197 + 198 + 199 - 200 We can rewrite it as: S = (1 + 2 + 3 + \cdots + 200) - 2(4 + 8 + \cdots + 200)

The sum of the first 200 integers: 1 + 2 + \cdots + 200 = \dfrac{200 \times 201}{2} = 20100 The second term is an arithmetic series of 4’s multiples: 4 + 8 + \cdots + 200 = 4(1 + 2 + \cdots + 50) = 4 \left( \dfrac{50 \times 51}{2} \right) = 4(1275) = 5100 Therefore, S = 20100 - 2(5100) = 20100 - 10200 = 9900

(B) 9900

x^2-12x+34=\pm2

Case 1: x^2-12x+34=2\Rightarrow x^2-12x+32=0

By using 31. Signs and Roots of Quadratics:
\Delta=12^2-4(32)=144-128=16>0 → there are 2 different real roots r_1,r_2.

By using 30. Vieta’s Formula:
r_1+r_2=-\frac{-12}{1}=12

Case 2: x^2-12x+34=-2\Rightarrow x^2-12x+36=0
(x-6)^2=0 → one root x=6.

The total sum of all real solutions is 12+6=18.

(C) 18

1+3+\cdots+(2m-1)=212+\big(2+4+\cdots+2n\big)

Use 25. Arithmetic Sequence and Series.

m^2 = 212 + n(n+1)

Use 6. Square of a Sum and 10. Difference of Squares.

m^2 = 212 + \Big(n+\tfrac12\Big)^2 - \tfrac14

4m^2 = 848 + (2n+1)^2 - 1

4m^2 - (2n+1)^2 = 847

(2m+2n+1)(2m-2n-1)=847

Since 2m+2n+1 and 2m-2n-1 are factors of 847.

(A) 255

Use 29. Integer and Fractional Parts.

Let x=k+r with k\in\mathbb{Z}, 0\le r<1. Then \lfloor x\rfloor=k.

Substitute:
k^2-3(k+r)+2=0\ \Rightarrow\ r=\dfrac{k^2-3k+2}{3}

Since 0\le r<1,
0\le \dfrac{k^2-3k+2}{3} < 1.

From the left inequality:
k^2-3k+2\ge 0 \Rightarrow (k-1)(k-2)\ge 0.
Using 31. Signs and Roots of Quadratics: k\in(-\infty,1]\cup[2,\infty).
Since k is an integer \Rightarrow k can be any integer.

From the right inequality:
\dfrac{k^2-3k+2}{3} < 1 \Rightarrow k^2-3k-1<0.
Use 21. Quadratic Formula: roots \dfrac{3\pm\sqrt{13}}{2}\approx-0....,\ 3.... .
By 31. Signs and Roots: -0....\lt k\lt 3....\Rightarrow k\in\{0,1,2,3\}.

Compute r=\dfrac{k^2-3k+2}{3} for each k:
k=0\Rightarrow r=\tfrac{2}{3}; k=1\Rightarrow r=0; k=2\Rightarrow r=0; k=3\Rightarrow r=\tfrac{2}{3}.
Thus x=k+r\in\left\{\tfrac{2}{3},\,1,\,2,\,\tfrac{11}{3}\right\} — four distinct values.

(B) 4

Add mn to both sides.

Use 6. Square of a sum.

(m+n)^2=m^2n^2+mn

By using 6. Square of a sum:
m^2n^2+mn=\left(mn+\tfrac12\right)^2-\tfrac14

From Hint 3 and Hint 4:
(m+n)^2=\left(mn+\tfrac12\right)^2-\tfrac14

Multiply both sides by 4:
(2m+2n)^2=(2mn+1)^2-1
(2mn+1)^2-(2m+2n)^2=1

By using 10. Difference of Squares:
\big(2mn+1+2m+2n\big)\big(2mn+1-2m-2n\big)=1

Case 1:
\begin{aligned} 2mn+1+2m+2n&=1\\ 2mn+1-2m-2n&=1 \end{aligned}
Adding: 4mn+2=2\Rightarrow mn=0\Rightarrow m=0\ \text{or}\ n=0
If m=0 \Rightarrow 2mn+1+2m+2n=1+2n=1\Rightarrow n=0
\Rightarrow (m,n)=(0,0)

Case 2:
\begin{aligned} 2mn+1+2m+2n&=-1\\ 2mn+1-2m-2n&=-1 \end{aligned}
\Rightarrow 4mn+2=-2\Rightarrow 4mn=-4\Rightarrow mn=-1
If m=1,\ n=-1 then 2mn+1+2m+2n=-2+1+2-2=-1 and 2mn+1-2m-2n=-2+1-2+2=-1.
Similarly, (m,n)=(-1,1).
Solutions: (m,n)=(1,-1),\,(-1,1).

(C) 3

Refer to 13. General difference of powers and 14. Sum of odd powers.

\(a-b \mid a^n-b^n,\quad n\in\mathbb{Z}^+\)

\(a+b \mid a^n+b^n,\quad n \text{ odd}\)

Choice A: \[ 2^{606}-1=(2^2)^{303}-1=4^{303}-1, \] so \(4-1=3 \mid 4^{303}-1\).

Choice B: \[ 2^{606}+1=4^{303}+1, \] so \(4+1=5 \mid 4^{303}+1\).

Choice D: \(2+1=3 \mid 2^{607}+1\).

Choice E: \(2+3=5 \mid 2^{607}+3^{607}\).

(C) \(2^{607}-1\)

Use 25. Arithmetic Sequence and Series.

\[ a_n = a_1 + (n-1)d = a_1 + 2(n-1) \]

\[ S_n = \frac{n}{2}(a_1+a_n) = \frac{n}{2}\big(2a_1+2(n-1)\big) \]

\[ S_{3n} = \frac{3n}{2}\big(2a_1+2(3n-1)\big) \]

Let \(a=a_1\). \[ \frac{S_{3n}}{S_n} = \frac{3\big(2a+2(3n-1)\big)}{2a+2(n-1)} \]

Since the ratio is independent of \(n\), \(\dfrac{S_3}{S_1}=\dfrac{S_6}{S_2}.\)

\[ \frac{3(2a+4)}{2a} = \frac{3(2a+10)}{2a+2} \]

\[ \frac{a+2}{a} = \frac{a+5}{a+1} \]

\[ a^2+5a = a^2+3a+2 \Rightarrow 2a=2 \Rightarrow a=1 \]

\[ S_{20} = \frac{20}{2}(a_1+a_{20}) =10\Big(1 + \big(1+19\cdot2\big)\Big) =400 \]

(D) 400

Use 5. Telescoping Series.

\[ \frac{k}{(k+1)!} = \frac{1}{k!} – \frac{1}{(k+1)!} \]

\[ \frac{1}{2!}+\frac{2}{3!}+\cdots+\frac{2021}{2022!} = \left(\frac{1}{1!}-\frac{1}{2!}\right) +\left(\frac{1}{2!}-\frac{1}{3!}\right) +\cdots +\left(\frac{1}{2021!}-\frac{1}{2022!}\right) = \frac{1}{1!}-\frac{1}{2022!} =1-\frac{1}{2022!} \] Therefore \(a=1\) and \(b=2022\), so \(a+b=2023\).

(D) 2023

Try small degrees for \(P(x)\). Use Remainder Theorem for quadratics.

If \(\deg P(x)=0\): \(P(x)=c\). Remainder on division by \(x^2+x+1\) would be \(c\), but must be \(x+2\). Impossible.

If \(\deg P(x)=1\): \(P(x)=ax+b\). From \(x^2+x+1\), remainder \(ax+b=x+2\Rightarrow a=1,b=2\). From \(x^2+1\), remainder \(ax+b=2x+1\Rightarrow a=2,b=1\). Contradiction.

If \(\deg P(x)=2\): \(P(x)=ax^2+bx+c\). Mod \(x^2+x+1=0\Rightarrow x^2=-x-1\): \[ ax^2+bx+c = a(-x-1)+bx+c = x(b-a)+(c-a). \] We need \(x(b-a)+(c-a)=x+2\Rightarrow b-a=1,\ c-a=2.\)
Mod \(x^2+1=0\Rightarrow x^2=-1\): \[ ax^2+bx+c = -a+bx+c = bx+(c-a). \] We need \(bx+(c-a)=2x+1\Rightarrow b=2,\ c-a=1.\) But \(c-a\) can’t be both 2 and 1. Impossible.

If \(\deg P(x)=3\): \(P(x)=ax^3+bx^2+cx+d\). In \(x^2+x+1=0\), we have \(x^2=-x-1\) and \(x^3=-x^2-x=1\). Then \[ P(x)\equiv a + b(-x-1) + cx + d = x(c-b) + (a-b+d). \] We want this to be \(x+2\), so \[ c-b=1,\quad a-b+d=2. \] In \(x^2+1=0\), we have \(x^2=-1,\ x^3=-x\). Then \[ P(x)\equiv a(-x)+b(-1)+cx+d = x(c-a)+(d-b). \] We want this \(=2x+1\), so \[ c-a=2,\quad d-b=1. \]

From the system \[ \begin{cases} c-b=1\\ a-b+d=2\\ c-a=2\\ d-b=1 \end{cases} \] Use \(d=b+1\) and \(c=a+2\). Then \(c-b=a+2-b=1 \Rightarrow a-b=-1.\) Also \(a-b+d = a-b+b+1 = 2 \Rightarrow a=1.\) Then \(b=a+1=2,\ c=a+2=3,\ d=3.\)

\(P(x)=x^3+2x^2+3x+3.\) Sum of squares of coefficients: \[ 1^2+2^2+3^2+3^2 = 23. \] (E) 23

Use 1. Sum of the first \(n\) integers and 24. Square of a sum (n terms).

\[ (1+2+\cdots+n)^2 = 1^2+2^2+\cdots+n^2 + 2\big(1\cdot2+1\cdot3+\cdots+(n-1)n\big) \]

\[ S_n = 1\cdot2 + 1\cdot3 + \cdots + (n-1)n = \frac{(1+2+\cdots+n)^2 – (1^2+2^2+\cdots+n^2)}{2} \]

Using known sums: \[ S_n = \frac{\left(\frac{n(n+1)}{2}\right)^2 – \frac{n(n+1)(2n+1)}{6}}{2} = \frac{n(n+1)(3n^2-n-2)}{24} = \frac{n(n+1)(3n+2)(n-1)}{24}. \]

For \(3\mid S_n\) we need \(n(n+1)(3n+2)(n-1)\) to be a multiple of \(9\). Since \(3n+2\) is not a multiple of 3, \((n-1)n(n+1)\) must be a multiple of 9. Among three consecutive integers exactly one is a multiple of 3, so that one must be a multiple of 9. Hence \(n \equiv 1,0,-1 \pmod{9}.\)

Since \(n\ge2\), the 10 least such values are \(8,9,10,17,18,19,26,27,28,35\).

(B) 197

Factor the expression: \[ x^{11} – 7x^{7} + x^{3} = x^{7}\!\Big(x^{4} – 7 + \frac{1}{x^{4}}\Big). \]

We’ll try to compute \(x^{4} + \dfrac{1}{x^{4}}\). To get that, first find \(x^{2} + \dfrac{1}{x^{2}}\). Use the square of a sum.

Calculate \(x^{2} + \dfrac{1}{x^{2}}\).

\[ x^{2} + \frac{1}{x^{2}} = \left(x + \frac{1}{x}\right)^{2} – 2 = (\sqrt{5})^{2} – 2 = 5 – 2 = 3. \]

Now calculate \(x^{4} + \dfrac{1}{x^{4}}\).

Square the result from Hint 4: \[ x^{4} + \frac{1}{x^{4}} = \left(x^{2} + \frac{1}{x^{2}}\right)^{2} – 2 = 3^{2} – 2 = 9 – 2 = 7. \]

Substitute back into the expression from Hint 1: \[ x^{11} – 7x^{7} + x^{3} = x^{7}\!\Big(\underbrace{x^{4} + \frac{1}{x^{4}}}_{=7} – 7\Big) = x^{7}(7 – 7) = x^{7} \cdot 0 = 0. \]

(B) 0

Rewrite as -(1^3+2^3+\cdots+18^3)+2(2^3+4^3+6^3+\cdots+18^3).

Use Identity 3. Sum of Cubes: 1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2.

Note 2^3+4^3+6^3+\cdots+18^3=8(1^3+2^3+\cdots+9^3).

(D) 3159

y-x=\pm 2y

Case 1: y-x=2y\ \Rightarrow\ y=-x

In Case 1:
y^3=x^2\ \Rightarrow\ (-x)^3=x^2\ \Rightarrow\ -x^3=x^2\ \Rightarrow\ x=0,-1
But x>0 (and y>0) so Case 1 gives no solution.

Case 2: y-x=-2y\ \Rightarrow\ x=3y

In Case 2:
y^3=x^2\ \Rightarrow\ y^3=(3y)^2\ \Rightarrow\ y^3=9y^2
\Rightarrow\ y=0,9; with y>0 we get y=9\Rightarrow x=27.

(D) 36

Use 28. AM–GM Inequality.

1+2a\ \ge\ 2\sqrt{2a}

2+2b\ \ge\ 2\sqrt{4b}=4\sqrt{b}

2a+b\ \ge\ 2\sqrt{2ab}

By multiplying Hints 2, 3, and 4:
(1+2a)(2+2b)(2a+b)\ \ge\ 32ab

By using 28. AM–GM Inequality, equality holds if and only if
1=2a,\qquad 2=2b,\qquad 2a=b

a=\tfrac12,\quad b=1

(B) 1

Use 32. Remainder of Polynomial Division.

Checking Choice (A):
\(x^2-x+1=0 \Rightarrow x^2=x-1\)
\(x^3=x^2\cdot x=(x-1)x=x^2-x=(x-1)-x=-1\)
\[ x^{2022}+x^{1011}+1=(x^3)^{674}+(x^3)^{337}+1 =(-1)^{674}+(-1)^{337}+1 =1\ne0 \]

Checking Choice (B):
\(x^2+x+1=0 \Rightarrow x^2=-x-1\)
\(x^3=-x^2-x=x+1-x=1\)
\[ x^{2022}+x^{1011}+1=(x^3)^{674}+(x^3)^{337}+1 =1+1+1=3\ne0 \]

Checking Choice (C):
\(x^4+1=0 \Rightarrow x^4=-1\)
\(x^{2022}=x^2(x^4)^{505}=-x^2,\quad x^{1011}=x^3(x^4)^{252}=x^3\)
\(x^{2022}+x^{1011}+1=-x^2+x^3+1\ne0\)

Checking Choice (D):
\(x^6-x^3+1=0 \Rightarrow x^6=x^3-1\)
\(x^9=x^6\cdot x^3=(x^3-1)x^3=x^6-x^3=(x^3-1)-x^3=-1\)
\[ x^{2022}+x^{1011}+1 =(x^9)^{224}x^6+(x^9)^{112}x^3+1 =(-1)^{224}x^6+(-1)^{112}x^3+1 =x^6+x^3+1 =2x^3\ne0 \]

Checking Choice (E):
\(x^6+x^3+1=0 \Rightarrow x^6=-x^3-1\)
\(x^9=x^6\cdot x^3=(-x^3-1)x^3=-x^6-x^3=-( -x^3-1)-x^3=1\)
\[ x^{2022}+x^{1011}+1 =(x^9)^{224}x^6+(x^9)^{112}x^3+1 =1^{224}x^6+1^{112}x^3+1 =x^6+x^3+1 =0 \]

(E) \(x^6+x^3+1\)

Use 20. Binomial Theorem.

Each term has the form \[ (x\sqrt[3]{2})^k(\sqrt{3}y)^{1000-k}, \quad 0 \le k \le 1000. \]

For \((\sqrt[3]{2})^k\) to be rational, \(k\) must be a multiple of 3.

For \((\sqrt{3})^{1000-k}\) to be rational, \(1000-k\) must be even, so \(k\) must be even.

So \(k\) must be a multiple of 6.

\[ 0 \le k = 6q \le 1000 \Rightarrow 0 \le q \le 166. \] There are 167 possible \(q\).

Final Answer: (C) 167

By 37. Trigonometric Transformations: \[ \cos\!\Big(\tfrac{\pi}{2}\sin x\Big) =\sin\!\Big(\tfrac{\pi}{2}-\tfrac{\pi}{2}\sin x\Big). \]

Use 36. General Solutions of Trigonometric Equations.

Case 1: \[ \tfrac{\pi}{2}\cos x = \tfrac{\pi}{2}-\tfrac{\pi}{2}\sin x+2\pi k \quad\Rightarrow\quad \sin x+\cos x=1+4k. \] Using identities: \[ \sqrt2\,\sin\!\big(x+\tfrac{\pi}{4}\big)=1 \Rightarrow \sin\!\big(x+\tfrac{\pi}{4}\big)=\tfrac{\sqrt2}{2}=\sin\tfrac{\pi}{4}. \]

By general solutions: \[ x+\tfrac{\pi}{4}=\tfrac{\pi}{4}+2\pi q \text{ or } \tfrac{3\pi}{4}+2\pi q \Rightarrow x=0,\tfrac{\pi}{2}. \]

Case 2: \[ \tfrac{\pi}{2}\cos x =-\Big(\tfrac{\pi}{2}-\tfrac{\pi}{2}\sin x\Big)+2\pi k \quad\Rightarrow\quad \cos x-\sin x=1+4(k-1). \] The feasible value is \(\cos x-\sin x=1.\) Then \[ -\sqrt2\,\sin\!\big(x-\tfrac{\pi}{4}\big)=1 \Rightarrow \sin\!\big(x-\tfrac{\pi}{4}\big)=-\tfrac{\sqrt2}{2} =\sin\!\big(-\tfrac{\pi}{4}\big). \]

General solutions: \[ x-\tfrac{\pi}{4}=-\tfrac{\pi}{4}+2\pi q \text{ or } \tfrac{5\pi}{4}+2\pi q \Rightarrow x=0 \text{ in }[0,\pi]. \] So the solutions in \([0,\pi]\) are \(x=0,\tfrac{\pi}{2}\).

Final Answer: (C) 2

Use 36. General Solutions of Trigonometric Equations.

\[ \sin x = \sin x^2 \Rightarrow x^2 = x + 360k \text{ or } x^2 = 180 – x + 360k, \quad k\in\mathbb{Z} \quad(\text{degrees}) \]

Case 1: \(x^2 = x + 360k\). \[ x^2 – x – 360k = 0 \Rightarrow x = \frac{1 \pm \sqrt{1 + 1440k}}{2}. \] Minimum \(x>1\) occurs at \(k=1\): \[ x = \frac{1 + \sqrt{1441}}{2} \approx \frac{1+38}{2} = 19.5. \]

Case 2: \(x^2 = 180 – x + 360k\). \[ x^2 + x – 360k – 180 = 0 \Rightarrow x = \frac{-1 \pm \sqrt{1440k + 721}}{2}. \] Minimum \(x>1\) is at \(k=0\): \[ x = \frac{-1 + \sqrt{721}}{2} \approx \frac{-1 + 27}{2} = 13. \] Round up: 13.
Final Answer: (B) 13

Use 30. Vieta’s Formula.

r_1 + r_2 + r_3 = -a
r_1 r_2 + r_1 r_3 + r_2 r_3 = b
r_1 r_2 r_3 = -6

r_1 r_2 r_3 = -6
r_1, r_2, and r_3 are integer roots.
WLOG, assume r_1 < r_2 < r_3

Use casework:

(A) 5

Use 30. Vieta’s Formula.

\[ r_1 + r_2 = -k,\quad r_1 r_2 = 36 \]

\(36 = 2^2\cdot 3^2\). Use 33. Number of Positive Divisors.

\(36\) has \((2+1)(2+1)=9\) positive divisors. Roots \(r_1\) and \(r_2\) are distinct, and \(k=-(r_1+r_2).\)

\(r_1 < r_2 \in \{\pm1,\pm2,\pm3,\pm4,\pm6,\pm9,\pm12,\pm18,\pm36\}\)
Pairs with product \(36\): \((1,36), (-1,-36), (2,18), (-2,-18), (3,12), (-3,-12), (4,9), (-4,-9)\)

\[ k = -(r_1 + r_2) = \pm 37,\ \pm 20,\ \pm 15,\ \pm 13 \]

(B) 8

\(c=\frac{2\pi}{11}\Rightarrow 11c=2\pi.\)

Use 37. Trigonometry Transformations.

\(\sin 6c=\sin(2\pi-6c)=-\sin(11c-6c)=-\sin 5c.\)

\(\sin 9c=\sin(11c-2c)=-\sin 2c.\)

\(\sin 12c=\sin(11c+c)=\sin c.\)

\(\sin 15c=\sin(11c+4c)=\sin 4c.\)

\[ \frac{\sin3c\cdot\sin6c\cdot\sin9c\cdot\sin12c\cdot\sin15c} {\sin c\cdot\sin2c\cdot\sin3c\cdot\sin4c\cdot\sin5c} = \frac{\sin3c\cdot(-\sin5c)\cdot(-\sin2c)\cdot\sin c\cdot\sin4c} {\sin c\cdot\sin2c\cdot\sin3c\cdot\sin4c\cdot\sin5c} =1. \]

(E) 1

Use Vieta’s Formula.

If \(r_1, r_2, r_3\) are the roots of \(f(x)\), then \[ r_1 + r_2 + r_3 = -a,\quad r_1r_2 + r_2r_3 + r_3r_1 = b,\quad r_1r_2r_3 = -c. \]

The polynomial \(g(x)\) has reciprocal roots, so \[ g(x) = (x – \tfrac{1}{r_1})(x – \tfrac{1}{r_2})(x – \tfrac{1}{r_3}). \]

Evaluate \(g(1)\): \[ g(1) = (1 – \tfrac{1}{r_1})(1 – \tfrac{1}{r_2})(1 – \tfrac{1}{r_3}) = \frac{(r_1-1)(r_2-1)(r_3-1)}{r_1r_2r_3} \] \[ = \frac{r_1r_2r_3 – (r_1r_2 + r_2r_3 + r_3r_1) + (r_1+r_2+r_3) – 1}{r_1r_2r_3}. \] Substitute from Vieta: \[ g(1) = \frac{-c – b – a – 1}{-c} = \frac{1 + a + b + c}{c}. \]

(A) \(\dfrac{1 + a + b + c}{c}\)

Use the Remainder Theorem for polynomial division modulo \(z^2+z+1\).

From \(z^2+z+1=0\) we get
z^2=-z-1 and hence
z^3=1 (for \(z\neq1\)).

Reduce the power:
z^{2021}=(z^3)^{673}\,z^2\equiv z^2\pmod{z^2+z+1}.
So
z^{2021}+1\equiv z^2+1\equiv(-z-1)+1=-z.
Therefore the remainder \(R(z)\) is \(-z\).

(A) -z

Use Identity 5. Telescoping Series.

(E) \frac{1}{22}

\binom{\binom{k}{2}}{2} = \frac{\binom{k}{2}\Big(\binom{k}{2}-1\Big)}{2} = \frac{\frac{k(k-1)}{2}\cdot\frac{k(k-1)-2}{2}}{2} \\ = \frac{1}{8}\,k(k-1)\big(k^2-k-2\big) = \frac{1}{8}\,k(k-1)(k-2)(k+1) \\ = \frac{1}{8}\cdot 4!\,\binom{k+1}{4} = 3\binom{k+1}{4}

Use 16. Hockey Stick Identity

3\binom{42}{5}=\frac{42\times41\times40\times39\times38}{5\times4\times2} =42\times41\times39\times38=(40+2)(40+1)(40-1)(40-2)

Use 10. Difference of Squares

(1600-4)(1600-1)

004

Use 24. Squares of a sum.

Expand and simplify it.

8x^2y^2z^2 - 4x^2y^2 - 4y^2z^2 - 4z^2x^2 + 2x^2 + 2y^2 + 2z^2 - 1 = 2023

Use 26. Factorization of three binomials.

(2x^2 - 1)(2y^2 - 1)(2z^2 - 1) = 2023

Prime factorize 2023.

(2x^2 - 1)(2y^2 - 1)(2z^2 - 1) = 7 \times 17^2

When a \in \mathbb{Z}^+, then 2a^2 - 1 \in \{1, 7, 17, 119, 289, 2023\}

a^2 = 1, 4, 9, 60, 145, 1012
\Rightarrow a = 1, 2, 3, \times, \times, \times

2x^2 - 1, 2y^2 - 1, 2z^2 - 1 \in \{1, 7, 17\}

2023 = 7 \times 17 \times 17

2x^2 - 1, 2y^2 - 1, 2z^2 - 1 is a permutation of \{7, 17, 17\}

(x, y, z) = (2, 3, 3), (3, 2, 3), (3, 3, 2)

WLOG, assume a \le b \le c.

We should prove 2(ab + bc + ca) + 4a^2 \ge a^2 + b^2 + c^2.

Use 24. Square of a Sum.

Add 2ab + 2ac + 2bc to both sides.

We should prove 4(ab + bc + ca + a^2) \ge (a + b + c)^2.

Use a + b + c = 4\sqrt[3]{abc}.

We should prove 4(4a\sqrt[3]{abc} + bc) \ge 16\sqrt[3]{a^2b^2c^2}.

Use 28. AM–GM Inequality.

By using 28. AM–GM Inequality, \frac{4a\sqrt[3]{abc} + bc}{2} \ge \sqrt{4a\sqrt[3]{abc} \cdot bc} = 2\sqrt[3]{a^2b^2c^2}.

Multiply both sides of Hint 9 by 8 to get the desired result.

a(a+1)(a+2)(a+3) = 4!\binom{a+3}{4}

Use Identity 16. Hockey Stick Identity.

3

Use Identity 22. Approaches to Similar Equations.

Add the two equations and find the value of xy.

Substitute xy into the first equation and find x - y.

Square x - y and use Identity 7. Square of a Difference.

\frac{9999}{64}

Use 27. Expansion (or factorization) of two binomials.

a^2 + 1 = a^2 + ab + ac + ca = (a + b)(a + c)

2(a + b)(a + c) = 3(b + a)(b + c) = 4(c + a)(c + b)

Let a + b = x,\ a + c = y,\ b + c = z.

2xy = 3yz = 4xz

y = \tfrac{3x}{4},\quad z = \tfrac{x}{2}

a + b = x,\ a + c = \tfrac{3x}{4},\ b + c = \tfrac{x}{2}

a = \tfrac{5x}{8},\quad b = \tfrac{3x}{8},\quad c = \tfrac{x}{8}

ab + ac + bc = 1

x = \tfrac{8}{\sqrt{23}}

a + b + c = \tfrac{x + y + z}{2}

\tfrac{9}{\sqrt{23}} = \tfrac{9\sqrt{23}}{23}

Use 6. Squares of a sum, 7. Squares of a difference and 10. Difference of squares

(n-3)^2 - 9 = (m+\tfrac{1}{2})^2 - \tfrac{1}{4} - 10

5 = (2m+1)^2 - (2n-6)^2 = (2m+2n-5)(2m-2n+7)

2m+2n-5 and 2m-2n+7 are factors of 5

(m,n) = (1,2),\ (1,4)

Use Identity 23: SFFT (Simon’s Favorite Factoring Trick).

(x-9)(y-9)=101

Since x and y are positive integers, \text{w.l.o.g. } x-9=1,\ y-9=101.

12200

Use 30. Vieta’s Formula.

\text{Volume} = (r_1 + 2)(r_2 + 2)(r_3 + 2)

By using 30. Vieta’s Formula:
\begin{aligned} r_1 + r_2 + r_3 &= \frac{39}{10}, \\ r_1r_2 + r_1r_3 + r_2r_3 &= \frac{29}{10}, \\ r_1r_2r_3 &= \frac{6}{10}. \end{aligned}

Expand:
(r_1 + 2)(r_2 + 2)(r_3 + 2) = r_1r_2r_3 + 2(r_1r_2 + r_2r_3 + r_3r_1) + 4(r_1 + r_2 + r_3) + 8. Substitute from Vieta’s results:
\frac{6}{10} + 2\left(\frac{29}{10}\right) + 4\left(\frac{39}{10}\right) + 8 = 30.

(D) 30

Use Identity 17. Vandermonde’s Identity.

y = \binom{40}{20}

(20!)^2 y = 40!

Find the number of factors of 5 in 40!.

Note that 25 contributes two factors of 5.

9