# SEAMO PAPER F 2019 [PROBLEM And SOLUTION]

Problem and Solution SEAMO 2019 paper E. Soal ini bersumber dari seamo-official.org

1. Find the integer part of

$$\frac{1}{\frac{1}{2015}+\frac{1}{2016}\frac{1}{2017}\frac{1}{2018}+\frac{1}{2019}}$$

(A) 400
(B) 401
(C) 402
(D) 403
(E) None of the above

Misalkan:

$$A=\frac{1}{\frac{1}{2015}+\frac{1}{2016}\frac{1}{2017}\frac{1}{2018}+\frac{1}{2019}}$$

$$\frac{1}{\frac{1}{2015}+\frac{1}{2015}\frac{1}{2015}\frac{1}{2015}+\frac{1}{2015}}<⌊A⌋<\frac{1}{\frac{1}{2019}+\frac{1}{2019}\frac{1}{2019}\frac{1}{2019}+\frac{1}{2019}}$$

$$\frac{1}{\frac{5}{2015}}<⌊A⌋<\frac{1}{\frac{5}{2019}}$$

$$\frac{2015}{5}<⌊A⌋<\frac{2019}{5}$$

$$403<⌊A⌋<403\frac{4}{5}$$

nilai bagian bulat dari A adalah 403

2. A box contains five cards, numbered 1, 2, 3, 4 and 5. Three cards are randomly selected without replacement from the box. What is the probability that “4” is the largest value selected?
(A) $$\frac{1}{5}$$
(B) $$\frac{3}{10}$$
(C) $$\frac{2}{5}$$
(D) $$\frac{1}{2}$$
(E) None of the above

Dari ketiga kartu, nomor kartu terbesar adalah 4 dan dua lainnya bernomor selain 5
Peluangnya adalah $$\frac{3\choose 2}{5\choose 3}=\frac{3}{10}$$

3. What fraction of all the 9-digit numbers formed using the digits 1 to 9, without repetition, is divisible by 36?
(A) $$\frac{1}{9}$$
(B) $$\frac{2}{9}$$
(C) $$\frac{1}{3}$$
(D) $$\frac{4}{9}$$
(E) None of the above

Habis dibagi 36, sama saja dengan habis dibagi 4 dan habis dibagi 9.
Bilangan 9 digit berbeda yang dibentuk dari angka {1, 2, 3, …,9} masing-masing digunakan sekali pasti habis dibagi 9 karena jumlah digitnya kelipatan 9.
Kita tinjau yang habis dibagi 4, Syarat habis dibagi 4 adalah 2 angka terakhir habis dibagi 4 yaitu $$\{12, 16, 24, 28, 32, 36, 48, 52, 56, 64, 68, 72, 76, 84, 92, 96\}$$ ada $$16$$ bilangan
Peluang habis dibagi 36 adalah
$$\frac{16×7!}{9!}=\frac{16×7!}{9×8×7!}=\frac{16}{9×8}=\frac{2}{9}$$

4. Find the remainder when

$$(x − 2)^{10} + (x − 3)^{20}$$

is divided by

$$x^2 − 5x + 6$$

(A) $$x + 1$$
(B) $$x$$
(C) $$1$$
(D) $$2$$
(E) None of the above

Misalkan sisanya adalah $$𝑎𝑥 + 𝑏$$
$$𝑥^2 − 5𝑥 + 6 = (𝑥 − 2)(𝑥 − 3)$$
Misalkan
$$𝑓(𝑥) = (𝑥 − 2)^{10} + (𝑥 − 3)^{20}$$
untuk $$𝑥 = 2$$
$$𝑓(2) = (2 − 2)^{10} + (2 − 3)^{20} = 2𝑎 + 𝑏$$
$$⇒ 1 = 2𝑎 + 𝑏$$
untuk $$𝑥 = 3$$
$$𝑓(3) = (3 − 2)^{10} + (3 − 3)^{20} = 2𝑎 + 𝑏$$
$$⇒ 1 = 3𝑎 + 𝑏$$
Karena $$2𝑎 + 𝑏 = 1$$ dan $$3𝑎 + 𝑏 = 1$$ maka $$𝑎 = 0, 𝑏 = 1$$
Jadi sisanya adalah $$𝑎𝑥 + 𝑏 = 0(𝑥) + 1 = 1$$

5. Find the coefficient of $$x^8$$ in

$$(x − 1)(x − 2)(x − 3)…(x − 10)$$

(A) 120
(B) 132
(C) 1200
(D) 1320
(E) None of the above

$$(𝑥 − 1)(𝑥 − 2) = 𝑥^2 − 3𝑥 + 2$$, koefisien dari $$𝑥^0$$ adalah $$2$$
$$(𝑥^2 − 3𝑥 + 2)(𝑥 − 3) = 𝑥^3 − 6𝑥^2 + 11𝑥 − 6$$ , koefisien dari $$𝑥^1$$ adalah $$11$$
$$(𝑥^3 − 6𝑥^2 + 11𝑥 − 6)(𝑥 − 4) = 𝑥^4 − 10𝑥^3 + 35𝑥^2 − 50𝑥 + 24$$, koefisien dari $$𝑥^2$$ adalah $$35$$
$$(𝑥^4 − 10𝑥^3 + 35𝑥^2 − 50𝑥 + 24)(𝑥 − 5) = 𝑥^5 − 15𝑥^4 + 85𝑥^3 − 225𝑥^2 + 274𝑥 − 12$$, koefisien dari $$𝑥^3$$ adalah $$85$$

Polanya $$2, 11, 35, 85,…$$
$$1.2 = 2$$
$$(1 + 2).3 + 1.2 = 11$$
$$(1 + 2 + 3)4 + (1 + 2)3 + 1. 2 = 35$$
$$(1 + 2 + 3 + 4)5 + (1 + 2 + 3)4 + (1 + 2)3 + 1. 2 = 85$$

$$(1 + 2 + 3 + ⋯ + 9)10 + (1 + 2 + 3 + ⋯ + 8)9 + (1 + 2 + 3 + ⋯ + 7)8 + (1 + 2 + 3 + ⋯ +6)7 + (1 + 2 + 3+ … + 5)6 + 85$$
$$= 45(10) + 36(9) + (28)(8) + 21(7) + 15(6) + 85$$
$$= 450 + 324 + 224 + 147 + 90 + 85$$
$$= 1320$$

6. Given that in quadrilateral $$ABCD$$,
i. $$∠ACB = ∠BAD = 105°$$
ii. $$∠ABC = ∠ADC = 45°$$

Which of the following is true?
(A) $$AB = CD$$
(B) $$AB = AD$$
(C) $$AB > CD$$
(D) $$AB < CD$$
(E) None of the above

7. Suppose

$$p+q=\frac{25}{4}$$

and

$$(1+\sqrt{p})(1+\sqrt{q})=\frac{15}{2}$$

Evaluate $$pq$$

(A) 1
(B) 2
(C) 3
(D) 4
(E) None of the above

$$𝑝 + 𝑞 =\frac{25}{4}⇒ (\sqrt{𝑝})^2+ (\sqrt{𝑞})^2 = (\sqrt{𝑝} + \sqrt{𝑞})^2− 2\sqrt{𝑝𝑞} =\frac{25}{4}$$

$$⇒ \sqrt{𝑝} + \sqrt{𝑞} = \sqrt{\frac{25}{4}+ 2\sqrt{𝑝𝑞}}$$

Selanjtunya
$$(1 + \sqrt{𝑝})(1 + \sqrt{𝑞}) =\frac{15}{2}$$

$$⇒1 + \sqrt{𝑝} + \sqrt{𝑞} + \sqrt {𝑝𝑞} = \frac{15}{2}$$

$$⇒\sqrt{\frac{25}{4}+ 2\sqrt{𝑝𝑞}} =\frac{13}{2}− \sqrt{𝑝𝑞}$$

$$⇒\frac{25}{4}+ 2\sqrt{𝑝𝑞} =\frac{169}{4}− 13\sqrt{𝑝𝑞} + 𝑝𝑞$$

$$⇒𝑝𝑞 − 15\sqrt{𝑝𝑞} + 36 = 0$$

$$⇒(\sqrt{𝑝𝑞} − 12)(\sqrt{𝑝𝑞} − 3) = 0$$

$$⇒\sqrt{𝑝𝑞} = 12$$ atau $$\sqrt{𝑝𝑞} = 3$$

$$⇒𝑝𝑞 = 144$$ atau $$𝑝𝑞 = 9$$

8. Gary wrote some numbers on each side of 3 cards, and laid them on a table, as shown.

Given that the sum of numbers on each card are equal and the numbers on the hidden sides are prime, find the sum of the three hidden numbers.
(A) 44
(B) 43
(C) 42
(D) 41
(E) None of the above

9. Find the number of positive integers x, such that $$4x^4 + 1$$ is prime.
(A) 0
(B) 1
(C) 2
(D) 3
(E) None of the above

10. Given that

$$2x=2019^{\frac{1}{2020}}-\frac{1}{2019^{\frac{1}{2020}}}$$

Evaluate

$$(\sqrt{1+x^2}-x)^{-2020}$$

(A) 0
(B) 1
(C) 2019
(D) 2020
(E) None of the above

Misalkan $$𝑎 = 2019^{\frac{1}{2020}}$$

$$2𝑥 = 𝑎 −\frac{1}{𝑎}$$
$$⇒4𝑥^2 = 𝑎^2 +\frac{1}{𝑎^2} − 2$$
$$⇒𝑥^2 =\frac{𝑎^2}{4}+\frac{1}{4𝑎^2} −\frac{1}{2}$$
$$⇒1 + 𝑥^2 = \frac{𝑎^2}{4}+\frac{1}{4𝑎^2} +\frac{1}{2}$$
$$⇒\sqrt{1 + 𝑥^2} =\sqrt{ \frac{𝑎^2}{4}+\frac{1}{4𝑎^2} +\frac{1}{2}}$$
$$⇒\sqrt{1 + 𝑥^2}=\sqrt(\frac{𝑎}{2}+\frac{1}{2𝑎})^2=\frac{𝑎}{2}+\frac{1}{2𝑎}$$
$$⇒\sqrt{1 + 𝑥^2} − 𝑥 = \frac{𝑎}{2}+\frac{1}{2𝑎}−(\frac{𝑎}{2}−\frac{1}{2𝑎})$$
$$⇒\sqrt{1 + 𝑥^2} − 𝑥 = (\frac{2}{2𝑎}) = (\frac{1}{𝑎}) = 𝑎^{−1}$$
$$⇒\sqrt{1 + 𝑥^2} − 𝑥 = (𝑎^{−1})^{−2020} = 𝑎^{2020} = (2019^{\frac{1}{2020}})^{2020}=2019$$

Pages ( 1 of 2 ): 1 2Next »