# SEAMO PAPER E 2019 [PROBLEM And SOLUTION]

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

1.Evaluate

$$\frac{4^8 + 4^8 +4^8 + 4^8}{3^8 + 3^8 +3^8} \times \frac{6^8 + 6^8 +6^8 + 6^8+6^8 + 6^8}{2^8 + 2^8}$$

(A) $$3^{16}$$
(B) $$3^{18}$$
(C) $$2^{16}$$
(D) $$2^{18}$$
(E) $$2^{20}$$

2. Find the number of pairs of positive integers α and β such that

$$α^2-β^2=361$$

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

$$𝑎^2 − 𝑏^2 = 361$$
$$(𝑎 + 𝑏)(𝑎 − 𝑏) = 361 = 19 × 19$$
Karena a dan bilangan bulat positif maka pasangan yang memenuhi hanya $$(𝑎 + 𝑏) = 361$$ dan $$(𝑎 − 𝑏) = 1$$, banyak pasangan $$(a,b)$$ yang memenuhi hanya $$1$$ pasang.

3. Given that $$(62)_n = 4 \times (14)_n$$ , find the
value of $$(36)_n$$ in base 10.
(A) 27
(B) 28
(C) 29
(D) 30
(E) None of the above

$$(62)_𝑛 = 4 × (14)_𝑛$$
$$6 × 𝑛 + 2 = 4 × (1 × 𝑛 + 4)$$
$$6𝑛 + 2 = 4𝑛 + 16$$
$$2𝑛 = 14$$
$$𝑛 = 7$$
Selanjutnya subtitusi nilai $$n$$ ke $$(36)_𝑛$$
$$(36)_7 = 3 × 7 + 6 = 27$$
Basis $$10$$
$$(27)_{10} = 2.10 + 7 = 27$$

4. Find the number of pairs of integers $$(m, n)$$ that satisfy the equation

$$m^3+6m^2+5m=21n^3+9n^2+3n+1$$

(A) 0
(B) 1
(C) 3
(D) 5
(E) None of the above

$$𝑚^3 + 6𝑚^2 + 5𝑚 = 21𝑛^3 + 9𝑛^2 + 3𝑛 + 1$$
$$21𝑛^3 + 9𝑛^2 + 3𝑛 + 1 ≡ 1\; 𝑚𝑜𝑑\; 3$$
Selanjutnya
$$𝑚^3 + 6𝑚^2 + 5𝑚 ≡ 1\; 𝑚𝑜𝑑\; 3$$
$$𝑚^3 + 5𝑚 ≡ 1\; 𝑚𝑜𝑑\; 3$$
$$𝑚(𝑚^2 + 2) ≡ 1\; 𝑚𝑜𝑑\; 3$$
$$𝑚$$ bukan kelipatan $$3$$,
Kita cek $$𝑚^2\; 𝑚𝑜𝑑\ 3$$
$$1^2 𝑚𝑜𝑑\; 3 = 1$$
$$2^2 𝑚𝑜𝑑\; 3 = 1$$
$$4^2 𝑚𝑜𝑑\; 3 = 1^2\; 𝑚𝑜𝑑\; 3 = 1$$
Dapat disimpulkan bahwa $$𝑚^2\; 𝑚𝑜𝑑\; 3 = 1$$, sehingga kita peroleh
$$𝑚(𝑚^2 + 2) ≡ 𝑚(1 + 2) ≡ 0\; 𝑚𝑜𝑑\; 3$$ (kontradiksi), tidak ada nilai m yang memenuhi.
Jadi banyaknya psangan $$(𝑚, 𝑛)$$ yang memenuhi adalah $$0$$

5. Find the digit in the ones place for

$$(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)(2^{64}+1)$$

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

6. Given that $$\log_{10} 2= m$$ and $$\log_{10} 3= n$$, find $$\log_{5} 24$$ in terms of $$m$$ and $$n$$.

(A) $$\frac{m+n}{1+m}$$
(B) $$\frac{2m+n}{1+m}$$
(C) $$\frac{2m+n}{1-m}$$
(D) $$\frac{3m+n}{1-m}$$
(E) None of the above

\begin{align} \log_5 {24} &=\frac{\log_{10}24}{\log_{10}5}\\ &=\frac{\log_{10} 8 + \log_{10} 3}{\log_{10}10 − \log_{10}2}\\ &=\frac{\log_{10}2^3 + \log_{10}3}{\log_{10}10 − \log_{10} 2}\\ &=\frac{3𝑚 + 𝑛}{1 − 𝑚}\\ \end{align}

7. Given that $$0≤θ≤\frac{π}{2}$$ and sin $$2θ=α$$, find $$\sin θ + \cos θ$$.

(A) $$\sqrt{a}$$
(B) $$\sqrt{a-1}$$
(C) $$\sqrt{a+1}$$
(D) $$\sqrt{a^2-a}$$
(E) None of the above

$$𝑚 = \sin 𝜃 + \cos 𝜃$$
$$𝑚^2 = (\sin 𝜃 + \cos 𝜃)^2 = \sin^2 𝜃 + \cos^2 𝜃 + 2 \sin 𝜃 \cos 𝜃 = 1 + \sin 2𝜃 = 1 + 𝑎$$
$$𝑚 = \sqrt{𝑎 + 1}$$

8. In ΔABC, AC=3AD, BC=3EC and BG : GF : FD = 3 : 1 : 2. Given that the area of ΔABC is 2700 cm³, find the area of ΔEFG in cm²

(A) 200
(B) 240
(C) 300
(D) 360
(E) None of the above

Misalkan $$[𝐺𝐹𝐸] = 𝑥$$
Karena $$𝐵𝐺 ∶ 𝐺𝐹 ∶ 𝐹𝐷 = 3 ∶ 1 ∶ 2$$ maka $$[𝐵𝐺𝐸] = 3𝑥$$ dan $$[𝐹𝐸𝐷] = 2𝑥$$
Karena $$𝐵𝐶 = 3𝐸𝐶$$ maka $$[EDC]=\frac{1}{2}[𝐵𝐸𝐷] =\frac{1}{2}(3𝑥 + 𝑥 + 2𝑥) = 3𝑥$$
Karena $$𝐴𝐶 = 3𝐴𝐷$$ maka $$[ABD]=\frac{1}{2}[𝐵𝐶𝐷] =\frac{1}{2}(9𝑥)=\frac{9}{2}𝑥$$
selanjutnya
$$[ABC] = [𝐵𝐶𝐷] + [𝐴𝐵𝐷] = 9𝑥 +\frac{9}{2}𝑥 = 2700 ⟹ 9𝑥 (\frac{3}{2}) = 2700 ⟹ 𝑥 = 200$$

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