HOTS Zone : Floor Function
Table of Contents
Tipe
No.
Jika(Catatan:
ALTERNATIF PENYELESAIAN
Misal x = ⌊x⌋ + p ⟶ ⌊x⌋ = x − p
\(\begin{aligned} \lfloor x\rfloor^2-2ax+a&=0\\ (x-p)^2-2ax+a&=0\\ x^2-2px+p^2-2ax+a&=0\\ x^2-(2p+2a)x+p^2+a&=0\\ \end{aligned}\)
2p + 2a = 51
\(p=\dfrac12\longrightarrow a=25\)
\(\begin{aligned} \lfloor x\rfloor^2-2ax+a&=0\\ (x-p)^2-2ax+a&=0\\ x^2-2px+p^2-2ax+a&=0\\ x^2-(2p+2a)x+p^2+a&=0\\ \end{aligned}\)
2p + 2a = 51
\(p=\dfrac12\longrightarrow a=25\)
Jadi, a = 25.
No.
Didefinisikan ⌊a⌋ menyatakan bilangan bulat terbesar yang lebih kecil atau sama dengan a. Perhatikan persamaan berikut \[\left\lfloor\dfrac{b}{1!}\right\rfloor+\left\lfloor\dfrac{b}{2!}\right\rfloor+\left\lfloor\dfrac{b}{3!}\right\rfloor+\cdots+\left\lfloor\dfrac{b}{2023!}\right\rfloor=999,\ b\in\mathbb{Z}\] Tentukan faktor prima terbesar dari b.ALTERNATIF PENYELESAIAN
\(\begin{array}{l}\left\lfloor\dfrac{7!}{1!}\right\rfloor\gt999\\[4pt]\left\lfloor\dfrac{6!}{1!}\right\rfloor+\left\lfloor\dfrac{6!}{2!}\right\rfloor=1080\gt999\\[4pt]\left\lfloor\dfrac{5!}{1!}\right\rfloor+\left\lfloor\dfrac{5!}{2!}\right\rfloor+\left\lfloor\dfrac{5!}{3!}\right\rfloor+\left\lfloor\dfrac{5!}{4!}\right\rfloor+\left\lfloor\dfrac{5!}{5!}\right\rfloor=206\end{array}\)
Didapat5! < b < 6! dan persamaan di atas dapat disederhanakan menjadi,\[\left\lfloor\dfrac{b}{1!}\right\rfloor+\left\lfloor\dfrac{b}{2!}\right\rfloor+\left\lfloor\dfrac{b}{3!}\right\rfloor+\left\lfloor\dfrac{b}{4!}\right\rfloor+\left\lfloor\dfrac{b}{5!}\right\rfloor=999\]misal \(f(x)=\left\lfloor\dfrac{x}{1!}\right\rfloor+\left\lfloor\dfrac{x}{2!}\right\rfloor+\left\lfloor\dfrac{x}{3!}\right\rfloor+\left\lfloor\dfrac{x}{4!}\right\rfloor+\left\lfloor\dfrac{x}{5!}\right\rfloor\)
Karena f(5!) = 206 makab = 4⋅5! + x1 dengan x1 < 5! .
999 − 4f(5!) = 175
\(f(4!)=\left\lfloor\dfrac{4!}{1!}\right\rfloor+\left\lfloor\dfrac{4!}{2!}\right\rfloor+\left\lfloor\dfrac{4!}{3!}\right\rfloor+\left\lfloor\dfrac{4!}{4!}\right\rfloor=41\)
Didapatx1 = 4⋅4! + x2 dengan x2 < 4! .
175 − 4f(4!) = 11
\(f(3!)=\left\lfloor\dfrac{3!}{1!}\right\rfloor+\left\lfloor\dfrac{3!}{2!}\right\rfloor+\left\lfloor\dfrac{3!}{3!}\right\rfloor=10\)
Didapatx2 = 3! + x3 dengan x3 < 3! .
11 − 10 = 1
f(1) = 1
Sehingga,
b = 4⋅5! + 4⋅4! + 3! + 1 = 583 = 53⋅11
Didapat
Karena f(5!) = 206 maka
999 − 4f(5!) = 175
\(f(4!)=\left\lfloor\dfrac{4!}{1!}\right\rfloor+\left\lfloor\dfrac{4!}{2!}\right\rfloor+\left\lfloor\dfrac{4!}{3!}\right\rfloor+\left\lfloor\dfrac{4!}{4!}\right\rfloor=41\)
Didapat
175 − 4f(4!) = 11
\(f(3!)=\left\lfloor\dfrac{3!}{1!}\right\rfloor+\left\lfloor\dfrac{3!}{2!}\right\rfloor+\left\lfloor\dfrac{3!}{3!}\right\rfloor=10\)
Didapat
11 − 10 = 1
f(1) = 1
Sehingga,
Jadi, faktor prima terbesar dari b adalah 53.
No.
Tentukan hasil dari penjumlahan \[\left\lfloor\dfrac79\right\rfloor+\left\lfloor\dfrac{14}9\right\rfloor+\left\lfloor\dfrac{21}9\right\rfloor+\cdots+\left\lfloor\dfrac{273}9\right\rfloor+\left\lfloor\dfrac{280}9\right\rfloor\]ALTERNATIF PENYELESAIAN
\begin{aligned}
\sum_{k=1}^9\left\lfloor\dfrac{7k}9\right\rfloor&=\left\lfloor\dfrac79\right\rfloor+\left\lfloor\dfrac{14}9\right\rfloor+\left\lfloor\dfrac{21}9\right\rfloor+\left\lfloor\dfrac{28}9\right\rfloor+\left\lfloor\dfrac{35}9\right\rfloor+\left\lfloor\dfrac{42}9\right\rfloor+\left\lfloor\dfrac{49}9\right\rfloor+\left\lfloor\dfrac{56}9\right\rfloor+\left\lfloor\dfrac{63}9\right\rfloor\\[3.8pt]
&=0+1+2+3+3+4+5+6+7\\
&=31
\end{aligned}
Perhatikan bahwa untuk m ≥ 0,
\begin{aligned}
\sum_{k=1}^9\left\lfloor\dfrac{7\cdot9m+7k}9\right\rfloor&=\sum_{k=1}^9\left\lfloor7m+\dfrac{7k}9\right\rfloor\\[3.8pt]
&=\sum_{k=1}^9\left(7m+\left\lfloor\dfrac{7k}9\right\rfloor\right)\\[3.8pt]
&=63m+31
\end{aligned}
Sehingga,
\begin{aligned}
\left\lfloor\dfrac79\right\rfloor+\left\lfloor\dfrac{14}9\right\rfloor+\left\lfloor\dfrac{21}9\right\rfloor+\cdots+\left\lfloor\dfrac{273}9\right\rfloor+\left\lfloor\dfrac{280}9\right\rfloor&=\sum_{m=0}^3(63m+31)+\left\lfloor\dfrac{7\cdot9\cdot4+7}9\right\rfloor+\left\lfloor\dfrac{7\cdot9\cdot4+14}9\right\rfloor+\left\lfloor\dfrac{7\cdot9\cdot4+21}9\right\rfloor+\left\lfloor\dfrac{7\cdot9\cdot4+28}9\right\rfloor\\[3.8pt]
&=63\cdot\dfrac42(0+3)+31\cdot4+28+(28+1)+(28+2)+(28+3)\\
&=\boxed{\boxed{620}}
\end{aligned}
Jadi, \(\left\lfloor\dfrac79\right\rfloor+\left\lfloor\dfrac{14}9\right\rfloor+\left\lfloor\dfrac{21}9\right\rfloor+\cdots+\left\lfloor\dfrac{273}9\right\rfloor+\left\lfloor\dfrac{280}9\right\rfloor=620\).
No.
Diberikan fungsi \(f(n)=1+\displaystyle\sum_{k=1}^n\left\lfloor{^3\negthinspace\log k}\right\rfloor\) yang terdefinisi pada bilangan asli n. Jika p adalah suatu bilangan sehingga- 440 ≤ p < 450
- 450 ≤ p < 460
- 460 ≤ p < 470
- 470 ≤ p < 480
- 480 ≤ p < 490
ALTERNATIF PENYELESAIAN
Misal a adalah bilangan bulat tak-negatif. Untuk 3a ≤ k ≤ 3a + 1 − 1,
⌊3log k⌋ = a
\(\begin{aligned} \displaystyle\sum_{k\ =\ 3^a}^{3^{a+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=a\left(3^{a+1}-1-3^a+1\right)\\ &=a\left(3\cdot3^a-3^a\right)\\ &=2a3^a \end{aligned}\)
\(\begin{aligned} \displaystyle\sum_{k\ =\ 1\ =\ 3^0}^{n\ =\ 3^{b+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2\displaystyle\sum_{a=0}^{b}a3^a\\ &=2\left(\dfrac{3-(b+1)3^{b+1}+b3^{b+2}}{(1-3)^2}\right)\\[4pt] &=2\left(\dfrac{3-(b+1)3^{b+1}+3b3^{b+1}}{4}\right)\\[4pt] &=\dfrac{3+(2b-1)3^{b+1}}2 \end{aligned}\)
Kita car nilai b sehingga \(\dfrac{3+(2b-1)3^{b+1}}2\) mendekati atau sama dengan 2022
\(\begin{aligned} \dfrac{3+(2b-1)3^{b+1}}2&\leq2022\\[4pt] 3+(2b-1)3^{b+1}&\leq4044\\ (2b-1)3^{b+1}&\leq4041\\ (2b-1)3^b&\leq1347 \end{aligned}\)
Jika b = 5,
(2(5) − 1)35 = 2187 > 1347
Jika b = 4,
(2(4) − 1)34 = 567 < 1347
didapat b = 4
\(\dfrac{3+(2(4)-1)3^{4+1}}2=852\)
\(\begin{aligned} f(p)&=2023\\ 1+\displaystyle\sum_{k=1}^p\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2023\\ \displaystyle\sum_{k=1}^{3^{4+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor+\displaystyle\sum_{k\ =\ 3^5\ =\ 243}^{p}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2022\\ 852+5(p-243+1)&=2022\\ p&=476 \end{aligned}\)
⌊3log k⌋ = a
\(\begin{aligned} \displaystyle\sum_{k\ =\ 3^a}^{3^{a+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=a\left(3^{a+1}-1-3^a+1\right)\\ &=a\left(3\cdot3^a-3^a\right)\\ &=2a3^a \end{aligned}\)
\(\begin{aligned} \displaystyle\sum_{k\ =\ 1\ =\ 3^0}^{n\ =\ 3^{b+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2\displaystyle\sum_{a=0}^{b}a3^a\\ &=2\left(\dfrac{3-(b+1)3^{b+1}+b3^{b+2}}{(1-3)^2}\right)\\[4pt] &=2\left(\dfrac{3-(b+1)3^{b+1}+3b3^{b+1}}{4}\right)\\[4pt] &=\dfrac{3+(2b-1)3^{b+1}}2 \end{aligned}\)
Kita car nilai b sehingga \(\dfrac{3+(2b-1)3^{b+1}}2\) mendekati atau sama dengan 2022
\(\begin{aligned} \dfrac{3+(2b-1)3^{b+1}}2&\leq2022\\[4pt] 3+(2b-1)3^{b+1}&\leq4044\\ (2b-1)3^{b+1}&\leq4041\\ (2b-1)3^b&\leq1347 \end{aligned}\)
Jika b = 5,
(2(5) − 1)35 = 2187 > 1347
Jika b = 4,
(2(4) − 1)34 = 567 < 1347
didapat b = 4
\(\dfrac{3+(2(4)-1)3^{4+1}}2=852\)
\(\begin{aligned} f(p)&=2023\\ 1+\displaystyle\sum_{k=1}^p\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2023\\ \displaystyle\sum_{k=1}^{3^{4+1}-1}\left\lfloor{^3\negthinspace\log k}\right\rfloor+\displaystyle\sum_{k\ =\ 3^5\ =\ 243}^{p}\left\lfloor{^3\negthinspace\log k}\right\rfloor&=2022\\ 852+5(p-243+1)&=2022\\ p&=476 \end{aligned}\)
Jadi, nilai p yang memenuhi berada pada interval 470 ≤ p < 480.
JAWAB: D
JAWAB: D
No.
Diberikan sistem persamaan berikut⌊x⌋ + ⌊y⌋ + y = 43,8
x + y − ⌊y⌋ = 18,4
Dengan x, y bilangan real. Nilai ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\lfloor x\rfloor+\lfloor y\rfloor+y&=43{,}8\\
\lfloor x\rfloor+\lfloor y\rfloor+\lfloor y\rfloor+\{y\}&=43{,}8\\
\lfloor x\rfloor+2\lfloor y\rfloor+\{y\}&=43{,}8
\end{aligned}\)
didapat,
⌊x⌋ + 2⌊y⌋ = 43, dan
{y} = 0,8
\(\begin{aligned} x+y-\left\lfloor y\right\rfloor&=18{,}4\\ \left\lfloor x\right\rfloor+\{x\}+\{y\}&=18{,}4\\ \left\lfloor x\right\rfloor+\{x\}+0{,}8&=18{,}4\\ \left\lfloor x\right\rfloor+\{x\}&=17{,}6 \end{aligned}\)
Didapat,
x = 17,6,
dan ⌊x⌋ = 17
\(\begin{aligned} \left\lfloor x\right\rfloor+2\left\lfloor y\right\rfloor&=43\\ 17+2\left\lfloor y\right\rfloor&=43\\ 2\left\lfloor y\right\rfloor&=26\\ \left\lfloor y\right\rfloor&=13\\ y&=13{,}8 \end{aligned}\)
\(\begin{aligned} 10x+10y&=10(17{,}6)+10(13{,}8)\\ &=176+138\\ &=\color{blue}\boxed{\boxed{\color{black}314}} \end{aligned}\)
didapat,
⌊x⌋ + 2⌊y⌋ = 43, dan
{y} = 0,8
\(\begin{aligned} x+y-\left\lfloor y\right\rfloor&=18{,}4\\ \left\lfloor x\right\rfloor+\{x\}+\{y\}&=18{,}4\\ \left\lfloor x\right\rfloor+\{x\}+0{,}8&=18{,}4\\ \left\lfloor x\right\rfloor+\{x\}&=17{,}6 \end{aligned}\)
Didapat,
x = 17,6,
dan ⌊x⌋ = 17
\(\begin{aligned} \left\lfloor x\right\rfloor+2\left\lfloor y\right\rfloor&=43\\ 17+2\left\lfloor y\right\rfloor&=43\\ 2\left\lfloor y\right\rfloor&=26\\ \left\lfloor y\right\rfloor&=13\\ y&=13{,}8 \end{aligned}\)
\(\begin{aligned} 10x+10y&=10(17{,}6)+10(13{,}8)\\ &=176+138\\ &=\color{blue}\boxed{\boxed{\color{black}314}} \end{aligned}\)
Jadi, nilai 10x + 10y yang memenuhi sistem persamaan di atas adalah 314.
No.
Jika penyelesaian dari persamaan berikut176x − 4⌊x⌋2 − 88{x}2 = 2023
dapat dinyatakan dalam bentuk $a-\dfrac1{\sqrt{b}}$ dengan a dan b bilangan bulat positif, maka ALTERNATIF PENYELESAIAN
\(\begin{aligned}
176x-4\lfloor x\rfloor^2-88\{x\}^2&=2023\\
176\left(\lfloor x\rfloor+\{x\}\right)-4\lfloor x\rfloor^2-88\{x\}^2&=2023\\
176\lfloor x\rfloor+176\{x\}-4\lfloor x\rfloor^2-88\{x\}^2&=2023\\
4\lfloor x\rfloor^2-176\lfloor x\rfloor+88\{x\}^2-176\{x\}&=-2023\\
4\left(\lfloor x\rfloor-22\right)^2+88\left(1-\{x\}\right)^2&=1
\end{aligned}\)
Perhatikan bahwa88(1 − {x})2 ≥ 0 .
Jika⌊x⌋ ≤ 21 atau ⌊x⌋ ≥ 23, maka 4(⌊x⌋ − 22)2 ≥ 4. Sehingga ⌊x⌋ = 22.
\(\begin{aligned} 88\left(1-\{x\}\right)^2&=1\\ 1-\{x\}&=\dfrac1{\sqrt{88}}\\[4pt] \{x\}&=1-\dfrac1{\sqrt{88}} \end{aligned}\)
$x=22+1-\dfrac1{\sqrt{88}}=23-\dfrac1{\sqrt{88}}$
23 + 88 =111
Perhatikan bahwa
Jika
\(\begin{aligned} 88\left(1-\{x\}\right)^2&=1\\ 1-\{x\}&=\dfrac1{\sqrt{88}}\\[4pt] \{x\}&=1-\dfrac1{\sqrt{88}} \end{aligned}\)
$x=22+1-\dfrac1{\sqrt{88}}=23-\dfrac1{\sqrt{88}}$
23 + 88 =
Jadi, a + b = 111 .
No.
Diberikan himpunan \[S=\left \{\left\lfloor\frac{1^3}{6211}\right\rfloor,\left\lfloor\frac{2^3}{6211}\right\rfloor,\left\lfloor\frac{3^3}{6211}\right\rfloor,\cdots,\left\lfloor\frac{6211^3}{6211}\right\rfloor\right\}\] Banyaknya anggota berbeda dari \(S\) adalah ....ALTERNATIF PENYELESAIAN
Perhatikan bahwa
$n^3-(n-1)^3=3n^2-3n+1$
\(\begin{aligned} 3n^2-3n+1&≥6211\\ 3n^203n-6210&≥0\\ n^2-n-2070&≥0\\ (n-46)(n+45)&≥0 \end{aligned}\)
Didapat \(n≥46\), artinya untuk \(n\geq46\) nilai \(\left\lfloor\frac{n^3}{6211}\right\rfloor\) dan \(\left\lfloor\frac{(n-1)^3}{6211}\right\rfloor\) selalu berbeda.
Untuk \(n=1\),
\(\left\lfloor\frac{1^3}{6211}\right\rfloor=0\)
Untuk \(n=45\),
\(\left\lfloor\frac{45^3}{6211}\right\rfloor=14\)
Ada 15 bilangan yang berbeda.
Dari \(n=46\) sampai \(n=6211\), ada \(6211-46+1=6166\) bilangan yang berbeda.
Sehingga total ada \(6166+15=6181\) bilangan yang berbeda.
$n^3-(n-1)^3=3n^2-3n+1$
\(\begin{aligned} 3n^2-3n+1&≥6211\\ 3n^203n-6210&≥0\\ n^2-n-2070&≥0\\ (n-46)(n+45)&≥0 \end{aligned}\)
Didapat \(n≥46\), artinya untuk \(n\geq46\) nilai \(\left\lfloor\frac{n^3}{6211}\right\rfloor\) dan \(\left\lfloor\frac{(n-1)^3}{6211}\right\rfloor\) selalu berbeda.
Untuk \(n=1\),
\(\left\lfloor\frac{1^3}{6211}\right\rfloor=0\)
Untuk \(n=45\),
\(\left\lfloor\frac{45^3}{6211}\right\rfloor=14\)
Ada 15 bilangan yang berbeda.
Dari \(n=46\) sampai \(n=6211\), ada \(6211-46+1=6166\) bilangan yang berbeda.
Sehingga total ada \(6166+15=6181\) bilangan yang berbeda.
Jadi, banyaknya anggota berbeda dari \(S\) adalah 6181.
No.
Diketahui k adalah bilangan bulat positif sehingga 5k adalah faktor dari- 40
- 35
- 32
- 30
- 28
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
125!-124\cdot124!-123\cdot123!&=124!\left(125-124\right)-123!\cdot123!\\
&=124!-123!\cdot123!\\
&=123!\left(124-123\right)\\
&=123!
\end{aligned}\)
\(\begin{aligned} k&=\left\lfloor\dfrac{123}5\right\rfloor+\left\lfloor\dfrac{123}{5^2}\right\rfloor\\[4pt] &=24+4\\ &=\color{blue}\boxed{\boxed{\color{black}28}} \end{aligned}\)
\(\begin{aligned} k&=\left\lfloor\dfrac{123}5\right\rfloor+\left\lfloor\dfrac{123}{5^2}\right\rfloor\\[4pt] &=24+4\\ &=\color{blue}\boxed{\boxed{\color{black}28}} \end{aligned}\)
Jadi, nilai terbesar k yang memenuhi adalah 28.
JAWAB: E
JAWAB: E
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