HOTS Zone : Pangkat (Eksponen) [2]

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} \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}

\(2=24^{\frac1x}\),   \(3=24^{\frac1y}\),   \(4=24^{\frac1z}\)

\(\begin{aligned} 24^{\frac1x+\frac1y+\frac1z}&=24^{\frac1x}\cdot24^{\frac1y}\cdot24^{\frac1z}\\ 24^{\frac1x+\frac1y+\frac1z}&=2\cdot3\cdot4\\ 24^{\frac1x+\frac1y+\frac1z}&=24\\ \dfrac1x+\dfrac1y+\dfrac1z&=\boxed{\boxed{1}} \end{aligned}\)

\(\begin{aligned} 8^n&=\left(2^3\right)^n\\ &=2^{3n} \end{aligned}\)

\(\begin{aligned} 44^{44}&=\left(2^2\cdot11\right)^{44}\\ &=2^{88}\cdot11^{44}\\ &=2^{3\cdot29+1}\cdot11^{44}\\ &=2^{3\cdot29}\cdot2\cdot11^{44} \end{aligned}\)

\(\begin{aligned} 2^{2002}\cdot5^{2003}&=2^{2002}\cdot5^{2002}\cdot5\\ &=(2\cdot5)^{2002}\cdot5\\ &=10^{2002}\cdot5\\ &=5\cdot10^{2002} \end{aligned}\)
Kita tahu bahwa 5 ⋅ 10²⁰⁰² berupa bilangan dengan 1 digit 5 di kiri, dan 2002 digit 0. Sehingga jumlah digit-digitnya adalah 5.

\(\begin{aligned} \dfrac{7^{777}}{5^{777}}\times\sqrt{\dfrac{5^{1554}+30^{1554}}{7^{1554}+42^{1554}}}&=\dfrac{7^{777}}{5^{777}}\times\sqrt{\dfrac{5^{1554}+5^{1554}\cdot6^{1554}}{7^{1554}+7^{1554}\cdot6^{1554}}}\\[4pt] &=\dfrac{7^{777}}{5^{777}}\times\sqrt{\dfrac{5^{1554}\left(1+6^{1554}\right)}{7^{1554}\left(1+6^{1554}\right)}}\\[4pt] &=\dfrac{7^{777}}{5^{777}}\times\dfrac{5^{777}}{7^{777}}\\ &=\color{blue}\boxed{\boxed{\color{black}1}} \end{aligned}\)

3(3^{n − 1} + 5^{n − 1}) < 3ⁿ + 5ⁿ < 5(3^{n − 1} + 5^{n − 1})
Sehingga,
\(\begin{aligned} 3^n+5^n&=4\left(3^{n-1}+5^{n-1}\right)\\ 3\cdot3^{n-1}+5\cdot5^{n-1}&=4\cdot3^{n-1}+4\cdot5^{n-1}\\ 5^{n-1}&=3^{n-1}\\ n-1&=0\\ n&=\color{blue}\boxed{\boxed{\color{black}1}} \end{aligned}\)

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.