# GUANGDONG-HONG KONG-MACAO GREATER BAY AREA MATHEMATICAL OLYMPIAD 2019 (MOCK EXAM) SECONDARY 8

Berikut ini soal dan solusi MOCK PAPER BBC 2019 Secondary 8, semoga bermanfaat

1. Find the remainder of $$2014^{504}÷7$$ .

\begin{align} 2014^{504}\; mod\; 7 &= 5^{504}\; mod\; 7\\ &= (5^3)^{168}\; mod\; 7\\ &= (125)^{168}\; mod\; 7\\ &= (−1)^{168}\; mod\; 7\\ &= 1\; mod\; 7\\ &= 1 \end{align}

2. Refer to the figure below, $$AB = AC. AD = DE = EF = FB = BC$$. Find the value of $$∠DEF$$.

misalkan $$∠BAC=a$$ karena $$AB = AC. AD = DE = EF = FB = BC$$ maka diperoleh sudut yang lain seperti gambar berikut

$$∠𝐴 + ∠𝐵 + ∠𝐶 = 180° ⟹ 𝑎° + 4𝑎° + 4𝑎° = 180° ⟹ 𝑎 = 20°$$
$$∠𝐷𝐸𝐹 = 180° − 4𝑎° = 180° − 4(20°) = 180° − 80° = 100°$$

3. According to the following equation, find the value of $$a+b+c+d$$

$$a+b+c=17$$
$$a+b+d=15$$
$$a+c+d=10$$
$$b+c+d=15$$

Jumlahkan keempat persamaan
$$3𝑎 + 3𝑏 + 3𝑐 + 3𝑑 = 57$$
$$𝑎 + 𝑏 + 𝑐 + 𝑑 = 19$$

4. Solve the equation $$𝑥^3 – x^2 – 4x – 6 = 0$$ .

$$𝑥^3 – x^2 – 4x – 6 = 0$$
$$⇒𝑥^2(𝑥 − 3) + 2𝑥^(𝑥 − 3) + 2(𝑥 − 3) = 0$$
$$⇒(𝑥 − 3)(𝑥2 + 2𝑥 + 2) = 0$$
Diperoleh $$𝑥 − 3 = 0$$ atau $$𝑥^2 + 2𝑥 + 2 = 0$$
Untuk $$𝑥 − 3 = 0 ⇒ 𝑥 = 3$$
Untuk $$𝑥^2 + 2𝑥 + 2 = 0$$ (tidak mempunyai solusi karena $$𝐷 < 0$$)

5. Factorize $$𝑥^2 + 3𝑥𝑦 + 2𝑦^2 − 6𝑥 − 7𝑦 + 5$$

$$(𝑥 + 2𝑦 − 5)(𝑥 + 𝑦 − 1)$$

6. Inside a class election, we have to choose 1 for chairperson, 3 for secretary and 1 for finance. If 30 students can take the above posts, how many post arrangements are there?

$${30\choose 1}{29\choose 3}{26\choose 1}= 2.850.120$$ cara

7. If $$x$$ and $$y$$ are positive integers, find the number of solutions of

$$\frac{1}{𝑥}+ \frac{1}{𝑦} =\frac{1}{8}$$

$$\frac{1}{𝑥}+ \frac{1}{𝑦} =\frac{1}{8}$$
$$\frac{𝑥 + 𝑦}{𝑥𝑦}=\frac{1}{8}$$
$$𝑥𝑦 = 8𝑥 + 8𝑦$$
$$𝑥𝑦 − 8𝑥 − 8𝑦 = 0$$
$$(𝑥 − 8)(𝑦 − 8) = 64$$
Jadi banyaknya pasangan $$(𝑥, 𝑦)$$ sama dengan banyaknya factor positif dari $$64$$ yaitu sebanyak $$7$$ pasang

8. For any positive integers $$n$$, it is known that $$n -5$$ and $$𝑛^2 + 10$$ are prime numbers. Find the value of $$n$$.

Karena $$𝑛^2 + 10 > 2$$ dan bilangan prima lebih dari dua selalu ganjil maka nilai $$𝑛$$ pasti ganjil.
Karena $$n$$ ganjil maka $$𝑛 − 5$$ genap, $$𝑛 − 5$$ bilangan prima, bilangan prima genap yang memenuhi hanya angka $$2$$, jadi nilai $$𝑛 = 7$$

9. Simplify $$\sqrt{3 + 2\sqrt 2} − \sqrt{3 − 2\sqrt 2}$$

$$\sqrt{3 + 2\sqrt 2} − \sqrt{3 − 2\sqrt 2}=\sqrt 2 + 1 − (\sqrt 2 − 1) = 2$$

10. From 1 to 100 to choose numbers such that any 2 numbers are not in an integral-multiple relationship, how many numbers can be chosen?

Bilangan yang diambil adalah bilangan prima $$\{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\}$$ jadi banyaknya bilangan yang dipilih ada 25 bilangan.

11. The perimeter of a rectangle with integral lengths is 2014. Find the minimum value of the area of this rectangle.

$$𝐾 = 2(𝑝 + 𝑙) = 2014$$
$$𝑝 + 𝑙 = 1007$$
Agar luas minimal maka nilai $$𝑝 = 1006$$ dan $$𝑙 = 1$$, jadi luasnya adalah $$1006 × 1 = 1006$$

12. Find the value of $$\frac{13577}{13581×13578−13579^2}$$

Misalkan $$𝑎 = 13579$$

$$\frac{13577}{13581×13578−13579^2}$$$$=\frac{𝑎−2}{(𝑎+2)(𝑎−1)−𝑎^2}$$$$=\frac{a-2}{a^2+a-2-a^2}$$$$=\frac{𝑎−2}{𝑎−2}=1$$

13. Refer to the figure below, $$ABCD$$ is a square. $$E$$ and $$F$$ lie on $$BC$$ and $$CD$$ respectively. It is
known that $$BE + FD =EF$$ . Find the value of $$EAF$$ .

Karena $$𝐵𝐸 + 𝐹𝐷 = 𝐸𝐹$$ maka $$𝐸𝐺 =𝐵𝐷$$ dan $$𝐺𝐹 = 𝐹𝐷$$, diperoleh $$Δ𝐴𝐵𝐸 ≅Δ𝐴𝐸𝐺$$ dan $$Δ𝐴𝐺𝐹 ≅ Δ𝐴𝐹𝐷$$
Karena $$Δ𝐴𝐵𝐸 ≅ Δ𝐴𝐸𝐺$$ dan $$Δ𝐴𝐺𝐹 ≅Δ𝐴𝐹𝐷$$ maka $$∠𝐵𝐴𝐸 = ∠𝐸𝐴𝐺$$ dan $$∠𝐺𝐴𝐹 = ∠𝐹𝐴𝐷$$
$$∠𝐵𝐴𝐸 + ∠𝐸𝐴𝐺 + ∠𝐺𝐴𝐹 + ∠𝐹𝐴𝐷 = 90°$$
$$2(∠𝐸𝐴𝐺 + ∠𝐺𝐴𝐹) = 90°$$
$$(∠𝐸𝐴𝐺 + ∠𝐺𝐴𝐹) = 45°$$
$$∠𝐸𝐴𝐹 = 45°$$

14. It is know that $$𝑥 +\frac{1}{𝑥}=3$$. Find the value of $$𝑥^3 − 𝑥^2 − 5𝑥 + 5$$.

$$𝑥 +\frac{1}{𝑥}= 3 ⇒ 𝑥^2 − 3𝑥 + 1 = 0$$
Selanjutnya

$$𝑥^3−𝑥^2−5𝑥+5$$
$$=𝑥(𝑥^2−3𝑥+1)+2𝑥^2−6𝑥+5$$

$$=2𝑥^2−6𝑥+5$$
$$=2(𝑥^2−3𝑥+1)+3$$
$$= 3$$

15. It is known that $$𝑎$$ dan $$𝑏$$ are positive integers and $$𝑎^2 − 𝑏^2 = 20140504$$. Find the maximum value of $$𝑏$$.

$$𝑎^2 − 𝑏^2 = 20140504$$
$$(𝑎 + 𝑏)(𝑎 − 𝑏) = 5035126 × 4$$
Diperoleh $$(𝑎 + 𝑏) = 5035126$$ dan $$𝑎 − 𝑏 = 4$$
Kurangkan kedua persamaan diperoleh
$$2𝑏 = 5035122 ⟹ 𝑏 = 2.517.561$$

16. A sequence of positive integers $$𝑎_1, 𝑎_2, 𝑎_3, …$$ satisfies the relation: $$𝑎_𝑛 = 𝑎_{𝑛−1} + 2𝑎_{𝑛−2}$$ and $$𝑛 ≥ 3$$. If $$𝑎_2 > 𝑎_1$$ and $$𝑎_4 = 11$$, what is the possible maximum value of $$𝑎_1$$?

Untuk $$𝑛 = 3$$
$$𝑎_3 = 𝑎_2 + 2𝑎_1$$
Untuk $$𝑛 = 4$$
$$𝑎_4 = 𝑎_3 + 2𝑎_2$$
$$11 = 𝑎_2 + 2𝑎_1 + 2𝑎_2 = 𝑎_2 + 4𝑎_1$$
Pasangan $$(𝑎_1, 𝑎_2)$$ yang memenuhi adalah $$(1, 7), (2, 3)$$
Jadi nilai maksimum $$𝑎_1$$ adalah $$3$$

17. Find the unit digit of $$11^2 + 13^3 + 15^3 + ⋯ + 199^3$$

Belum tersedia

18. Inside a right-anged triangle, $$sin\;θ=\frac{7}{25}$$. Find the value of $$Tan\;θ$$

Belum tersedia

19. If $$𝑥 = 1,2,3, . . . ,200$$ and substitute all $$x$$ into $$𝑦 = | 𝑥 − 100|$$ . Find the sum of all values of $$y$$.

$$|1 − 100| + |2 − 100| + |3 − 100| + ⋯ + |100 − 100| + |101 − 100 + ⋯ + |200 − 100|$$
$$= (99 + 98 + ⋯ + 1) + 0 + (1 + 2 + 3 + ⋯ + 100)$$
$$= 4950 + 5050$$
$$= 10000$$

20. Solve the equation of $$(𝑥2 − 3𝑥 + 2)^{𝑥−1} = 0$$

$$𝑥^2 − 3𝑥 + 2 = 0$$
$$(𝑥 − 2)(𝑥 − 1) = 0$$
$$𝑥 = 2$$ atau $$𝑥 = 1$$(tidak memenuhi karena $$\frac{0}{0} ≠ 0$$)
Jadi solusinya adalah $$𝑥 = 2$$

21. Find the least positive integral solution of $$11x ≡7 (mod\;19)$$

$$11x ≡7 (mod\;19)$$
$$22x ≡14 (mod\;19)$$
$$3x ≡14 (mod\;19)$$
$$18x ≡84 (mod\;19)$$
$$-x ≡8 (mod\;19)$$
$$x ≡-8 (mod\;19)$$
$$x ≡11 (mod\;19)$$
$$𝑥 = 11$$