HOTS Zone : Pangkat (Eksponen) [2]
Table of Contents
Tipe
No.
Agar bilangan yang berbentukALTERNATIF PENYELESAIAN
Misal 213 + 210 + 2n = p2
\(\begin{aligned} 2^{10}\left(2^3+1\right)+2^n&=p^2\\ 2^{10}\left(9\right)+2^n&=p^2\\ 2^{10}\cdot3^2+2^n&=p^2\\ 2^n&=p^2-2^{10}\cdot3^2\\ 2^{a+b}&=p^2-\left(2^5\cdot3\right)^2&{\color{red}\text{misal }n=a+b}\\ 2^a\cdot 2^b&=\left(p+2^5\cdot3\right)\left(p-2^5\cdot3\right) \end{aligned}\)
\(\begin{aligned} 2^a&=p+2^5\cdot3\\ 2^b&=p-2^5\cdot3&-\\\hline 2^a-2^b&=2\cdot2^5\cdot3\\ 2^b\left(2^{a-b}-1\right)&=2^6\cdot3 \end{aligned}\)
b = 6
a − b = 2 ⟶ a = 8
n = a + b = 8 + 6 = 14
\(\begin{aligned} 2^{10}\left(2^3+1\right)+2^n&=p^2\\ 2^{10}\left(9\right)+2^n&=p^2\\ 2^{10}\cdot3^2+2^n&=p^2\\ 2^n&=p^2-2^{10}\cdot3^2\\ 2^{a+b}&=p^2-\left(2^5\cdot3\right)^2&{\color{red}\text{misal }n=a+b}\\ 2^a\cdot 2^b&=\left(p+2^5\cdot3\right)\left(p-2^5\cdot3\right) \end{aligned}\)
\(\begin{aligned} 2^a&=p+2^5\cdot3\\ 2^b&=p-2^5\cdot3&-\\\hline 2^a-2^b&=2\cdot2^5\cdot3\\ 2^b\left(2^{a-b}-1\right)&=2^6\cdot3 \end{aligned}\)
b = 6
a − b = 2 ⟶ a = 8
n = a + b = 8 + 6 = 14
Jadi, n = 14.
No.
Bilangan \(\dfrac{\left(2^4\right)^8}{\left(4^8\right)^2}\) sama dengan- \(\dfrac14\)
- \(\dfrac12\)
- 1
- 2
- 8
ALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{\left(2^4\right)^8}{\left(4^8\right)^2}&=\dfrac{2^{32}}{\left(2^2\right)^{16}}\\[4pt]
&=\dfrac{2^{32}}{2^{32}}\\
&=\boxed{\boxed{1}}
\end{aligned}
Jadi, bilangan \(\dfrac{\left(2^4\right)^8}{\left(4^8\right)^2}\) sama dengan 1.
JAWAB: C
JAWAB: C
No.
\(\dfrac1x+\dfrac1y+\dfrac1z=\) ....
ALTERNATIF PENYELESAIAN
\(2=24^{\frac1x}\), \(3=24^{\frac1y}\), \(4=24^{\frac1z}\)
\(\begin{aligned} 24^{\frac1x+\frac1y+\frac1z}&=24^{\frac1x}\cdot24^{\frac1y}\cdot24^{\frac1z}\\ 24^{\frac1x+\frac1y+\frac1z}&=2\cdot3\cdot4\\ 24^{\frac1x+\frac1y+\frac1z}&=24\\ \dfrac1x+\dfrac1y+\dfrac1z&=\boxed{\boxed{1}} \end{aligned}\)
\(\begin{aligned} 24^{\frac1x+\frac1y+\frac1z}&=24^{\frac1x}\cdot24^{\frac1y}\cdot24^{\frac1z}\\ 24^{\frac1x+\frac1y+\frac1z}&=2\cdot3\cdot4\\ 24^{\frac1x+\frac1y+\frac1z}&=24\\ \dfrac1x+\dfrac1y+\dfrac1z&=\boxed{\boxed{1}} \end{aligned}\)
Jadi, \(\dfrac1x+\dfrac1y+\dfrac1z=1\).
No.
Bilangan n terbesar sehingga 8n membagi 4444 adalah ....- 8
- 22
- 29
- 44
- 88
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
8^n&=\left(2^3\right)^n\\
&=2^{3n}
\end{aligned}\)
\(\begin{aligned} 44^{44}&=\left(2^2\cdot11\right)^{44}\\ &=2^{88}\cdot11^{44}\\ &=2^{3\cdot29+1}\cdot11^{44}\\ &=2^{3\cdot29}\cdot2\cdot11^{44} \end{aligned}\)
\(\begin{aligned} 44^{44}&=\left(2^2\cdot11\right)^{44}\\ &=2^{88}\cdot11^{44}\\ &=2^{3\cdot29+1}\cdot11^{44}\\ &=2^{3\cdot29}\cdot2\cdot11^{44} \end{aligned}\)
Jadi, bilangan n terbesar sehingga 8n membagi 4444 adalah 29.
JAWAB: C
JAWAB: C
No.
Berapakah jumlah digit-digit bilanganALTERNATIF PENYELESAIAN
\(\begin{aligned}
2^{2002}\cdot5^{2003}&=2^{2002}\cdot5^{2002}\cdot5\\
&=(2\cdot5)^{2002}\cdot5\\
&=10^{2002}\cdot5\\
&=5\cdot10^{2002}
\end{aligned}\)
Kita tahu bahwa5 ⋅ 102002 berupa bilangan dengan 1 digit 5 di kiri, dan 2002 digit 0. Sehingga jumlah digit-digitnya adalah 5.
Kita tahu bahwa
Jadi, jumlah digit-digit bilangan 22002 ⋅ 52003 adalah 5.
No.
MisalkanALTERNATIF PENYELESAIAN
\(\begin{aligned}
A&=(-1)^{-1}\\
&=\dfrac1{-1}\\[4pt]
&=-1
\end{aligned}\)
\(\begin{aligned} B&=(-1)^1\\ &=-1 \end{aligned}\)
\(\begin{aligned} C&=1^{-1}\\ &=1 \end{aligned}\)
\(\begin{aligned} B&=(-1)^1\\ &=-1 \end{aligned}\)
\(\begin{aligned} C&=1^{-1}\\ &=1 \end{aligned}\)
\(\begin{aligned}
A+B+C&=-1+(-1)+1\\
&=\color{blue}\boxed{\boxed{\color{black}-1}}
\end{aligned}\)
Jadi, A + B + C = −1 .
No.
$$\dfrac{7^{777}}{5^{777}}\times\sqrt{\dfrac{5^{1554}+30^{1554}}{7^{1554}+42^{1554}}}=....$$ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\dfrac{7^{777}}{5^{777}}\times\sqrt{\dfrac{5^{1554}+30^{1554}}{7^{1554}+42^{1554}}}&=\dfrac{7^{777}}{5^{777}}\times\sqrt{\dfrac{5^{1554}+5^{1554}\cdot6^{1554}}{7^{1554}+7^{1554}\cdot6^{1554}}}\\[4pt]
&=\dfrac{7^{777}}{5^{777}}\times\sqrt{\dfrac{5^{1554}\left(1+6^{1554}\right)}{7^{1554}\left(1+6^{1554}\right)}}\\[4pt]
&=\dfrac{7^{777}}{5^{777}}\times\dfrac{5^{777}}{7^{777}}\\
&=\color{blue}\boxed{\boxed{\color{black}1}}
\end{aligned}\)
Jadi, $\dfrac{7^{777}}{5^{777}}\times\sqrt{\dfrac{5^{1554}+30^{1554}}{7^{1554}+42^{1554}}}=1$.
No.
Carilah semua bilangan asli n sedemikian sehingga3n − 1 + 5n − 1 | 3n + 5n
ALTERNATIF PENYELESAIAN
Sehingga,
\(\begin{aligned} 3^n+5^n&=4\left(3^{n-1}+5^{n-1}\right)\\ 3\cdot3^{n-1}+5\cdot5^{n-1}&=4\cdot3^{n-1}+4\cdot5^{n-1}\\ 5^{n-1}&=3^{n-1}\\ n-1&=0\\ n&=\color{blue}\boxed{\boxed{\color{black}1}} \end{aligned}\)
Jadi, n = 1.
No.
Misalkan N bilangan yang memenuhi persamaanJika faktorisasi prima dari N adalah
- 44
- 50
- 49
- 79
ALTERNATIF PENYELESAIAN
Karena kita menghitung hasil penjumlahan semua pangkat primanya, sama seperti kita menghitung berapakah banyaknya prima (belum tentu berbeda) yang perlu dikalikan untuk menghasilkan N. Tinjau bahwa faktorisasi prima dari 2,3,5,7 berisikan 1 bilangan prima, faktorisasi prima dari 4,6, dan 9 berisikan 2 bilangan prima, dan faktorisasi prima 8 berisikan 3 bilangan prima. Maka, banyaknya prima yang diperlukan adalah
(2 + 3 + 5 + 7) × 1 + (4 + 6 + 9) × 2 + (8) × 3 = 17 + 38 + 24 = 79.
Jadi, a + b + c + d = 79.
JAWAB: D
JAWAB: D
No.
Hasil dari $\dfrac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+\cdots}{1^{-3}+3^{-3}+5^{-3}+7^{-3}+\cdots}=$ ....- $\dfrac78$
- $\dfrac87$
- $\dfrac98$
- $\dfrac{15}{16}$
- $\dfrac{16}{15}$
ALTERNATIF PENYELESAIAN
Misal S = 1−3 + 2−3 + 3−3 + 4−3 + ⋯
\(\begin{aligned} \dfrac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+\cdots}{1^{-3}+3^{-3}+5^{-3}+7^{-3}+\cdots}&=\dfrac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+\cdots}{\left(1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+6^{-3}+7^{-3}+\cdots\right)-\left(2^{-3}+4^{-3}+6^{-3}+\cdots\right)}\\[4pt] &=\dfrac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+\cdots}{\left(1^{-3}+2^{-3}+3^{-3}+4^{-3}+\cdots\right)-2^{-3}\left(1^{-3}+2^{-3}+3^{-3}+\cdots\right)}\\[4pt] &=\dfrac{S}{S-2^{-3}S}\\[4pt] &=\dfrac{S}{S-\dfrac18S}\\[4pt] &=\dfrac{S}{\dfrac78S}\\[4pt] &=\boxed{\boxed{\dfrac87}} \end{aligned}\)
\(\begin{aligned} \dfrac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+\cdots}{1^{-3}+3^{-3}+5^{-3}+7^{-3}+\cdots}&=\dfrac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+\cdots}{\left(1^{-3}+2^{-3}+3^{-3}+4^{-3}+5^{-3}+6^{-3}+7^{-3}+\cdots\right)-\left(2^{-3}+4^{-3}+6^{-3}+\cdots\right)}\\[4pt] &=\dfrac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+\cdots}{\left(1^{-3}+2^{-3}+3^{-3}+4^{-3}+\cdots\right)-2^{-3}\left(1^{-3}+2^{-3}+3^{-3}+\cdots\right)}\\[4pt] &=\dfrac{S}{S-2^{-3}S}\\[4pt] &=\dfrac{S}{S-\dfrac18S}\\[4pt] &=\dfrac{S}{\dfrac78S}\\[4pt] &=\boxed{\boxed{\dfrac87}} \end{aligned}\)
Jadi, hasil dari $\dfrac{1^{-3}+2^{-3}+3^{-3}+4^{-3}+\cdots}{1^{-3}+3^{-3}+5^{-3}+7^{-3}+\cdots}=\dfrac87$.
JAWAB: B
JAWAB: B
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