# Problems And Solutions SEAMO PAPER E 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 E. Soal ini bersumber dari seamo-official.org

1. Find the value of the product

$$(1 +\frac{1}{2})(1 +\frac{1}{2^2})(1 +\frac{1}{2^4}) … (1 +\frac{1}{2^{1024}})$$

(A) $$2 −\frac{1}{2^{2047}}$$
(B) $$2 −\frac{1}{2^{2048}}$$
(C) $$1 −\frac{1}{2^{2048}}$$
(D) $$2 −\frac{1}{2^{1024}}$$
(E) None of the above

$$(1 +\frac{1}{2})(1 +\frac{1}{2^2})(1 +\frac{1}{2^4}) … (1 +\frac{1}{2^{1024}})$$

$$\frac{1}{2}𝐴 = (1 −\frac{1}{2})(1 +\frac{1}{2})(1 +\frac{1}{2^2})(1 +\frac{1}{2^4}) … (1 +\frac{1}{2^{1024}})$$

$$\frac{1}{2}𝐴 = (1 −\frac{1}{2^2})(1 +\frac{1}{2^2})(1 +\frac{1}{2^4}) … (1 +\frac{1}{2^{1024}})$$

$$\frac{1}{2}𝐴 = (1 −\frac{1}{2^4})(1 +\frac{1}{2^4}) … (1 +\frac{1}{2^{1024}})$$

$$\frac{1}{2}𝐴 = (1 −\frac{1}{2^{1024}})(1 +\frac{1}{2^{1024}})$$

$$\frac{1}{2}𝐴 = (1 −\frac{1}{2^{2048}})$$

$$𝐴 = 2 −\frac{1}{2^{2047}}$$

2. How many positive integers $$𝑛$$ satisfy the condition?

$$3^{200} < 𝑛^{100} < (123𝑛)^{50}$$

(A) 112
(B) 113
(C) 114
(D) 115
(E) None of the above

Untuk $$3^{200} < 𝑛^{100} ⟹ 9^{100} < 𝑛^{100} ⟹ 𝑛 > 9$$
Untuk $$𝑛^{100} < (123𝑛)^{50} ⟹ (𝑛^2)^{50} < (123𝑛)^{50} ⟹ 𝑛^2 < 123𝑛 ⟹ 𝑛 < 123$$
Diperoleh batasan nilai $$n$$ adalah $$9 < 𝑛 < 123$$, nilai $$n$$ yang memenuhi adalah $$\{10, 11, 12, …, 122\}$$ banyaknya ada $$113$$ bilangan

3. Denote $$𝑎_𝑛$$ by the last two digits of $$6^𝑛$$, for all positive integers $$𝑛$$.
For example, $$𝑎_1 = 06, 𝑎_2 = 36, 𝑎_3 = 16…$$
Evaluate the last two digits of the sum
$$𝑎_1 + 𝑎_2 + 𝑎_3 + ⋯ + 𝑎_{2021}$$.
(A) 22
(B) 24
(C) 26
(D) 28
(E) None of the above

$$𝑎_1 = 6^1\; 𝑚𝑜𝑑\; 100 ≡ 06$$
$$𝑎_2 = 6^2\; 𝑚𝑜𝑑\; 100 ≡ 36$$
$$𝑎_3 = 6^3\; 𝑚𝑜𝑑\; 100 ≡ 16$$
$$𝑎_4 = 6^4\; 𝑚𝑜𝑑\; 100 ≡ 96$$
$$𝑎_5 = 6^5\; 𝑚𝑜𝑑\; 100 ≡ 76$$
$$𝑎_6 = 6^6\; 𝑚𝑜𝑑\; 100 ≡ 56$$
$$𝑎_7 = 6^7\; 𝑚𝑜𝑑\; 100 ≡ 36$$
$$𝑎_8 = 6^8\; 𝑚𝑜𝑑\; 100 ≡ 16$$

Dari pola di atas diperoleh $$(𝑎_2 + 𝑎_3 + 𝑎_4 + 𝑎_5 + 𝑎_6) = (𝑎_7 + 𝑎_8 + 𝑎_9 + 𝑎_{10} + 𝑎_{11}) = ⋯$$
Karena dari $$𝑎_2$$ sampai dengan $$𝑎_{2021}$$ ada $$2020$$ suku maka
$$(𝑎_2 + 𝑎_3 + 𝑎_4 + 𝑎_5 + 𝑎_6) = (𝑎_7 + 𝑎_8 + 𝑎_9 + 𝑎_{10} + 𝑎_{11}) = ⋯ = (𝑎_{2017} + 𝑎_{2018} + 𝑎_{2019} + 𝑎_{2020} + 𝑎_{2021}) = (36 + 16 + 96 + 76 + 56)\;𝑚𝑜𝑑\; 100 ≡ 80$$
$$𝑎_1 + 𝑎_2 + 𝑎_3 + ⋯ + 𝑎_{2021} = 𝑎_1 + (𝑎_2 + 𝑎_3 + 𝑎_4 + 𝑎_5 + 𝑎_6) + (𝑎_7 + 𝑎_8 + 𝑎_9 + 𝑎_{10} + 𝑎_{11}) + ⋯ +(𝑎_{2017} + 𝑎_{2018} + 𝑎_{2019} + 𝑎_{2020} + 𝑎_{2021})$$
$$=(06 +\frac{2020}{5}(80))\; 𝑚𝑜𝑑\; 100 ≡ 26$$

4. Suppose $$𝐴𝐵𝐶𝐷$$ is a rectangle. $$𝑋$$ and $$𝑌$$ are points on $$𝐵𝐶$$ and $$𝐶𝐷$$, respectively, such that areas of $$Δ𝐴𝐵𝑋 , Δ𝐶𝑋𝑌$$ and $$Δ𝐴𝑌𝐷$$ are $$3, 4$$ and $$5$$, respectively. Evaluate the area of $$Δ𝐴𝑋𝑌$$.
(A) 6
(B) 7
(C) 8
(D) 9
(E) None of the above

5. Find the value of $$𝑎$$ for which

$$\sqrt{𝑎 + \sqrt{𝑎 + \sqrt{𝑎+. . .}}} = 7$$

(A) 40
(B) 42
(C) 45
(D) 49
(E) None of the above

$$\sqrt{𝑎 + \sqrt{𝑎 + \sqrt{𝑎+. . .}}} = 7$$
$$\sqrt{𝑎 + 7} = 7$$
$$𝑎 + 7 = 49$$
$$𝑎 = 42$$

Pages ( 1 of 5 ): 1 23 ... 5Next »