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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