# Problems And Solutions SEAMO PAPER F 2021

The Southeast Asia Mathematical Olympiad (SEAMO) is an international Math Olympiad competition that originated in Singapore and was founded by Mr. Terry Chew in 2016 in 8 Southeast Asian Countries. Since then, it is growing its popularity around the world. In 2019 it was recognized by 18 countries. In 2020 total number of participating countries increased to 22, including students from Indonesia, Brazil, China, Newzealand, and Taiwan students enrolled in SEAMO 2022-23.

Problem and Solution SEAMO 2021 paper F. Soal ini bersumber dari seamo-official.org

1. How many positive integers $$𝑛$$ are there such that

$$\frac{1}{3}<\frac{𝑛}{11}<\frac{1}{2}$$

(A) 0
(B) 1
(C) 2
(D) 3
(E) None of the above

$$\frac{1}{3}<\frac{𝑛}{11}<\frac{1}{2}$$
$$\frac{22}{66}<\frac{6𝑛}{66}<\frac{33}{66}$$
$$22 < 6𝑛 < 33$$
$$\frac{22}{6}< 𝑛 <\frac{33}{6}$$
$$3\frac{4}{6}< 𝑛 < 5\frac{3}{6}$$
Nilai $$n$$ bilangan bulat yang memenuhi adalah 4 dan 5 yaitu sebanyak 2 buah

2. Suppose $$𝑎_𝑛$$ is an arithmetic sequence with the first term $$𝑎_1 = 1$$ and common difference $$𝑑 = 8$$. Evaluate the sum

$$𝑆 =\frac{1}{\sqrt{𝑎_1} + \sqrt{𝑎_2}}+\frac{1}{\sqrt{𝑎_2} + \sqrt{𝑎_3}}+\frac{1}{\sqrt{𝑎_3} + \sqrt{𝑎_4}}+ ⋯ +\frac{1}{\sqrt{𝑎_{253}} + \sqrt{𝑎_{254}}}$$

(A) 3
(B) 4
(C) $$\frac{9}{2}$$
(D) $$\frac{11}{2}$$
(E) None of the above

Misalkan $$x_n=\frac{1}{\sqrt{𝑎_n} + \sqrt{𝑎_{n+1}}}$$
$$x_n=\frac{1}{\sqrt{𝑎_n} + \sqrt{𝑎_{n+1}}}×\frac{\sqrt{𝑎_n} – \sqrt{𝑎_{n+1}}}{\sqrt{𝑎_n} – \sqrt{𝑎_{n+1}}}=\frac{\sqrt{𝑎_n} – \sqrt{𝑎_{n+1}}}{a_n – a_{n+1}}=\frac{\sqrt{𝑎_n} – \sqrt{𝑎_{n+1}}}{a_1+(n-1)d – (a_1 +(n+1-1)d)}=\frac{\sqrt{𝑎_n} – \sqrt{𝑎_{n+1}}}{-d}=\frac{\sqrt{𝑎_{n+1}} – \sqrt{𝑎_n}}{d}$$

$$𝑆 =\frac{1}{\sqrt{𝑎_1} + \sqrt{𝑎_2}}+\frac{1}{\sqrt{𝑎_2} + \sqrt{𝑎_3}}+\frac{1}{\sqrt{𝑎_3} + \sqrt{𝑎_4}}+ ⋯ +\frac{1}{\sqrt{𝑎_{253}} + \sqrt{𝑎_{254}}}$$

$$=x_1 + x_2 + x_3 + … +x_{252}+x_{253}$$

$$=\frac{\sqrt{𝑎_2} – \sqrt{𝑎_1}}{d}+\frac{\sqrt{𝑎_3} – \sqrt{𝑎_2}}{d}+ +\frac{\sqrt{𝑎_4} – \sqrt{𝑎_3}}{d} + … + \frac{\sqrt{𝑎_{253}} – \sqrt{𝑎_{252}}}{d}+ \frac{\sqrt{𝑎_{254}} – \sqrt{𝑎_{253}}}{d}$$

$$=\frac{\sqrt{𝑎_{254}} – \sqrt{𝑎_1}}{d}$$

$$=\frac{\sqrt{𝑎_1+253(8)} – \sqrt{𝑎_1}}{8}$$

$$=\frac{\sqrt{1+253(8)} – \sqrt{1}}{8}$$

$$=\frac{\sqrt{2025} – \sqrt{1}}{8}=\frac{45-1}{8}=\frac{11}{2}$$

3. Suppose the probability that it rains on Saturday is $$\frac{1}{3}$$ and the probability that it rains on Sunday is $$\frac{1}{3}$$. What is the probability that it rains on weekends?
(A) $$\frac{1}{3}$$
(B) $$\frac{5}{9}$$
(C) $$\frac{2}{3}$$
(D) $$\frac{7}{9}$$
(E) None of the above

Peluang hujan di hari Sabtu $$P(S) =\frac{1}{3}$$
Peluang tidak hujan di hari Sabtu $$P(S’)=\frac{2}{3}$$
Peluang hujan di hari Minggu $$P(M) =\frac{1}{3}$$
Peluang tidak hujan di hari Minggu $$P(M’)=\frac{2}{3}$$

Peluang hujan di waktu weekends sama saja peluang hujan di hari Sabtu atau Minggu yaitu

$$P(S)P(M’) + P(S’)P(M) + P(S)P(M)=(\frac{1}{3})(\frac{2}{3})+(\frac{2}{3})(\frac{1}{3})+(\frac{1}{3})(\frac{1}{3})=(\frac{2}{9})+(\frac{2}{9})+(\frac{1}{9})=\frac{5}{9}$$

4. Let $$α$$ and $$β$$ be the roots of the quadratic equation $$𝑥^2 − 2𝑏𝑥 + 𝑏 = 1$$ .
What is the minimum value of $$|α − β|$$ ?
(A) $$\sqrt 2$$
(B) $$\sqrt 3$$
(C) $$2$$
(D) $$3$$
(E) None of the above

Berdasarkan dalil vieta
$$𝛼 + 𝛽 = 2𝑏$$
$$𝛼𝛽 = 𝑏 − 1$$

$$(𝛼 + 𝛽)^2 = 𝛼^2 + 𝛽^2 + 2𝛼𝛽$$
$$⇒(2𝑏)^2 = 𝛼^2 + 𝛽^2 + 2(𝑏 − 1)$$
$$⇒𝛼^2 + 𝛽^2 = 4𝑏^2 − 2(𝑏 − 1)$$

$$(𝛼 − 𝛽)^2 = 𝛼^2 + 𝛽^2 − 2𝛼𝛽$$
$$𝛼 − 𝛽 = \sqrt{𝛼^2 + 𝛽^2 − 2𝛼𝛽}$$
$$= \sqrt{4𝑏^2 − 2(𝑏 − 1) − 2(𝑏 − 1)}$$
$$= \sqrt{4𝑏^2 − 4𝑏 + 4}$$
$$= \sqrt{(2𝑏 − 1)2 + 3}$$
Nilai minimum dicapai ketika $$2𝑏 − 1 = 0$$ yaitu $$\sqrt 3$$

5. A test consists of 100 questions. A student gets 4, -1 or 0 marks if he answers a question correctly, wrongly, or left it blank, respectively. How many different total marks of the test are there? (A total mark can be negative).
(A) 250
(B) 475
(C) 500
(D) 501
(E) None of the above

Skor yang tidak mungkin ada dari 1 sampai dengan 400 adalah 399, 398, 397, 394, 393 dan 389. Skor yang mungkin dari 1 sampai dengan 400 ada sebanyak 394. Skor berbeda yang mungkin dari 0 sampai dengan -100 ada sebanayak 101.
Jadi total skor berbeda yang mungkin adalah 394 + 101 = 495

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