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No. 41

Berapa banyak n dari 1 hingga 2021 yang mengakibatkan n² + 3 dan n + 4 tidak relatif prima?

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ALTERNATIF PENYELESAIAN

Misal FPB dari n² + 3 dan n + 4 adalah k ≠ 1.
Misal n + 4 = ka, dan n² + 3 = kb, dengan a dan b bilangan asli.

n = ka − 4

$\begin{array}{rl}{n}^2+3& =kb\\ (ka-4{)}^2+3& =kb\\ k^2{a}^2-8ka+16+3& =kb\\ k^2{a}^2-8ka+19& =kb\end{array}$
kita lihat bahwa k | 19, sehingga didapat k = 19.

$\begin{array}{rcccl}1& <& n& <& 2021\\ 1& <& 19a-4& <& 2021\\ 5& <& 19a& <& 2025\\ {\displaystyle \frac{5}{19}}& <& a& <& {\displaystyle \frac{2025}{19}}\\ 1& \le & a& \le & 106\end{array}$
ada 106 nilai a yang memenuhi, sehingga ada 106 juga nilai n yang memenuhi.

Jadi, ada 106 nilai n dari 1 hingga 2021 yang mengakibatkan n² + 3 dan n + 4 tidak relatif prima.

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No. 42

Diberikan bilangan real positif x, y, z yang memenuhi sistem persamaan berikut
xyz = 1
$x+\dfrac1z=5$
$y+\dfrac1x=29$
$z+\dfrac1y=\dfrac{m}n$
Dengan FPB(m, n) = 1. Nilai dari m + n adalah ....

Penugasan GMOM 1 Februari 2024

ALTERNATIF PENYELESAIAN

Misal $p=z+\dfrac1y=\dfrac{m}n$

$\begin{array}{rl}(x+{\displaystyle \frac{1}{z}})(y+{\displaystyle \frac{1}{x}})(z+{\displaystyle \frac{1}{y}})& =x+{\displaystyle \frac{1}{z}}+y+{\displaystyle \frac{1}{x}}+z+{\displaystyle \frac{1}{y}}+xyz+{\displaystyle \frac{1}{xyz}}\\ \left(5\right)\left(29\right)\left(p\right)& =5+29+p+1+{\displaystyle \frac{1}{1}}\\ 145p& =p+36\\ 144p& =36\\ p& ={\displaystyle \frac{36}{144}}\\ {\displaystyle \frac{m}{n}}& ={\displaystyle \frac{1}{4}}\end{array}$
m = 1
n = 4

m + n = 1 + 4 = 5

Jadi, m + n = 5

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No. 43

Diketahui a, b ∈ ℝ yang memenuhi $$a+{\displaystyle \frac{1}{a+2015}}=b-4030+{\displaystyle \frac{1}{b-2015}}$$ dan |a − b| > 5000. Tentukan nilai dari $${\displaystyle \frac{ab}{2015}}-a+b.$$ (Notasi ℝ menyatakan himpunan semua bilangan real)

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ALTERNATIF PENYELESAIAN

$\begin{array}{rl}a+{\displaystyle \frac{1}{a+2015}}& =b-4030+{\displaystyle \frac{1}{b-2015}}\\ a+2015+{\displaystyle \frac{1}{a+2015}}& =b-2015+{\displaystyle \frac{1}{b-2015}}\end{array}$
Misal x = a + 2015, dan y = b − 2015. Jika x = y, maka |a − b| = 4030, sehingga x ≠ y.
$\begin{array}{rl}x+{\displaystyle \frac{1}{x}}& =y+{\displaystyle \frac{1}{y}}\\ x-y& ={\displaystyle \frac{1}{y}}-{\displaystyle \frac{1}{x}}\\ x-y& ={\displaystyle \frac{x-y}{xy}}\\ xy& =1\\ (a+2015)(b-2015)& =1\\ ab-2015a+2015b-4060225& =1\\ ab-2015a+2015b& =4060226\\ {\displaystyle \frac{ab}{2015}}-a+b& =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black{\displaystyle \frac{4060226}{2015}}}}}}\end{array}$

Jadi, $\dfrac{ab}{2015}-a+b=\dfrac{4060226}{2015}$

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No. 44

Hitunglah $${\displaystyle \frac{({10}^4+2^6)({18}^4+2^6)({26}^4+2^6)({34}^4+2^6)({42}^4+2^6)}{(6^4+2^6)({14}^4+2^6)({22}^4+2^6)({30}^4+2^6)({38}^4+2^6)}}$$

UG-HSMT 2008

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{x}^4+2^6& =x^4+4\cdot 2^4\\ & =(x^2+2\cdot 2^2-2\cdot 2\cdot x)(x^2+2\cdot 2^2+2\cdot 2\cdot x)\\ & =(x(x-4)+8)(x(x+4)+8)\end{array}$

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$\begin{array}{rl}{\displaystyle \frac{({10}^4+2^6)({18}^4+2^6)({26}^4+2^6)({34}^4+2^6)({42}^4+2^6)}{(6^4+2^6)({14}^4+2^6)({22}^4+2^6)({30}^4+2^6)({38}^4+2^6)}}& ={\displaystyle \frac{(10(10-4)+8)(10(10+4)+8)(18(18-4)+8)(18(18+4)+8)\cdots (42(42-4)+8)(42(42+4)+8)}{(6(6-4)+8)(6(6+4)+8)(14(14-4)+8)(14(14+4)+8)\cdots (38(38-4)+8)(38(38+4)+8)}}\\ & ={\displaystyle \frac{(10(6)+8)(10(14)+8)(18(14)+8)(18(22)+8)\cdots (42(38)+8)(42(46)+8)}{(6(2)+8)(6(10)+8)(14(10)+8)(14(18)+8)\cdots (38(34)+8)(38(42)+8)}}\\ & ={\displaystyle \frac{1940}{20}}\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black97}}}}\end{array}$

Jadi, $\dfrac{\left(10^4+2^6\right)\left(18^4+2^6\right)\left(26^4+2^6\right)\left(34^4+2^6\right)\left(42^4+2^6\right)}{\left(6^4+2^6\right)\left(14^4+2^6\right)\left(22^4+2^6\right)\left(30^4+2^6\right)\left(38^4+2^6\right)}=97$.

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No. 45

Tentukan banyak bilangan asli n sedemikian sehingga $$\frac{n^{2024}}{n+1}+\frac{5}{n^2+n}$$ bilangan asli

KTOM Agustus 2023

ALTERNATIF PENYELESAIAN

${\displaystyle \frac{n^{2024}}{n+1}}+{\displaystyle \frac{5}{n^2+n}}={\displaystyle \frac{n^{2025}+5}{n^2+n}}$

$\begin{array}{rl}{n}^{2025}+5& =n^{2025}+n^{2024}-n^{2024}+5\\ & =-n^{2024}+5& (\mathrm{mod}{n}^2+n)\\ & =-n^{2024}-n^{2023}+n^{2023}+5& (\mathrm{mod}{n}^2+n)\\ & =n^{2023}+5& (\mathrm{mod}{n}^2+n)\\ & =n+5& (\mathrm{mod}{n}^2+n)\end{array}$

agar ${\displaystyle \frac{n+5}{n^2+n}}$ bilangan asli, maka
$\begin{array}{rl}{n}^2+n& \le n+5\\ n^2& \le 5\end{array}$
n = 1 atau n = 2

• n = 1

  $\begin{array}{rl}{\displaystyle \frac{n+5}{n^2+n}}& ={\displaystyle \frac{1+5}{1^2+1}}\\ & =3\end{array}$

• n = 2

  $\begin{array}{rl}{\displaystyle \frac{n+5}{n^2+n}}& ={\displaystyle \frac{2+5}{2^2+2}}\\ & ={\displaystyle \frac{7}{6}}\end{array}$
  Bukan bilangan bulat

Jadi, banyak bilangan asli n sedemikian sehingga $$\frac{n^{2024}}{n+1}+\frac{5}{n^2+n}$$ bilangan asli ada 1 bilangan.

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No. 46

Pertidaksamaan berikut berlaku untuk semua bilangan bulat positif n. $$\sqrt{n+1}-\sqrt{n}<{\displaystyle \frac{1}{\sqrt{4n+1}}}<\sqrt{n}-\sqrt{n-1}$$ Berapa nilai bilangan bulat terbesar yang kurang dari $\displaystyle\sum_{n=1}^{24}\dfrac1{\sqrt{4n+1}}$?

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ALTERNATIF PENYELESAIAN

$$\begin{array}{rcccl}{\displaystyle \sum _{n=1}^{24}(\sqrt{n+1}-\sqrt{n})}& <& {\displaystyle \sum _{n=1}^{24}{\displaystyle \frac{1}{\sqrt{4n+1}}}}& <& {\displaystyle \sum _{n=1}^{24}(\sqrt{n}-\sqrt{n-1})}\\ \sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\cdots +\sqrt{25}-\sqrt{24}& <& {\displaystyle \sum _{n=1}^{24}{\displaystyle \frac{1}{\sqrt{4n+1}}}}& <& \sqrt{1}-\sqrt{0}+\sqrt{2}-\sqrt{1}+\cdots +\sqrt{24}-\sqrt{23}\\ -\sqrt{1}+\sqrt{25}& <& {\displaystyle \sum _{n=1}^{24}{\displaystyle \frac{1}{\sqrt{4n+1}}}}& <& -\sqrt{0}+\sqrt{24}\\ 4& <& {\displaystyle \sum _{n=1}^{24}{\displaystyle \frac{1}{\sqrt{4n+1}}}}& <& \sqrt{24}\\ 4& <& {\displaystyle \sum _{n=1}^{24}{\displaystyle \frac{1}{\sqrt{4n+1}}}}& <& 4,...\end{array}$$

Jadi, nilai bilangan bulat terbesar yang kurang dari $\displaystyle\sum_{n=1}^{24}\dfrac1{\sqrt{4n+1}}$ adalah 4.

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No. 47

Diberikan bilangan real x, y yang memenuhi persamaan x + y = 4 dan xy = −2. Nilai dari $$x+{\displaystyle \frac{x^2}{y^2}}+{\displaystyle \frac{y^2}{x^2}}+y$$ adalah ....

Evaluasi GMOM Februari 2024

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}x+{\displaystyle \frac{x^2}{y^2}}+{\displaystyle \frac{y^2}{x^2}}+y& =x+y+{\displaystyle \frac{x^4+y^4}{x^2{y}^2}}\\ & =x+y+{\displaystyle \frac{(x+y{)}^4-4xy(x+y{)}^2+2(xy{)}^2}{(xy{)}^2}}\\ & =4+{\displaystyle \frac{4^4-4(-2)(4{)}^2+2(-2{)}^2}{(-2{)}^2}}\\ & =4+{\displaystyle \frac{256+128+8}{4}}\\ & =4+{\displaystyle \frac{392}{4}}\\ & =4+98\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black102}}}}\end{array}$

Jadi, nilai dari $$x+{\displaystyle \frac{x^2}{y^2}}+{\displaystyle \frac{y^2}{x^2}}+y$$ adalah 102.

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No. 48

Diketahui a dan b dua bilangan real nonnegatif yang memenuhi a + 2b = 15. Nilai terbesar yang mungkin dari 3a + 5b adalah ....

1. 49
2. 45

3. 51
4. 41

KTOM Simulasi OSK SMP 2024

ALTERNATIF PENYELESAIAN

a = 15 − 2b

$\begin{array}{rl}3a+5b& =3(15-2b)+5b\\ & =45-6b+5b\\ & =45-b\end{array}$
Nilai 45 − b akan maksimum ketika nilai b minimum, yaitu b = 0.
45 − 0 = 45

Jadi, nilai terbesar yang mungkin dari 3a + 5b adalah 45.
JAWAB: B

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No. 49

Misalkan x, y bilangan real positif dengan x > y. Jika diketahui bahwa $x^2+y^2=\left(\dfrac{545}{272}\right)xy$, maka $\dfrac{x+y}{x-y}$ adalah ....

OSK SMA 2024

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}{\left({\displaystyle \frac{x+y}{x-y}}\right)}^2& ={\displaystyle \frac{x^2+y^2+2xy}{x^2+y^2-2xy}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{545}{272}}\right)xy+2xy}{\left({\displaystyle \frac{545}{272}}\right)xy-2xy}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{1089}{272}}\right)xy}{\left({\displaystyle \frac{1}{272}}\right)xy}}\\ & =1089\\ {\displaystyle \frac{x+y}{x-y}}& =\sqrt{1089}\\ & =\text{color}blue\boxed{{\displaystyle \boxed{{\displaystyle \text{color}black33}}}}\end{array}$

Jadi, $\dfrac{x+y}{x-y}=33$.

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No. 50

Jika $x+y+3\sqrt{x+y}=18$ dan $x-y-2\sqrt{x-y}=15$, maka xy = ....

Diktat Pembinaan Olimpiade Matematika SMA

ALTERNATIF PENYELESAIAN

$\begin{array}{rl}x+y+3\sqrt{x+y}& =18\\ {\left(\sqrt{x+y}\right)}^2+3\sqrt{x+y}-18& =0\\ (\sqrt{x+y}+6)(\sqrt{x+y}-3)& =0\\ \sqrt{x+y}& =3\\ x+y& =9\end{array}$

$\begin{array}{rl}x-y-2\sqrt{x-y}& =15\\ {\left(\sqrt{x-y}\right)}^2-2\sqrt{x-y}-15& =0\\ (\sqrt{x-y}+3)(\sqrt{x-y}-5)& =0\\ \sqrt{x-y}& =5\\ x-y& =25\end{array}$

$\begin{array}{rl}4xy& ={(x+y)}^2-{(x-y)}^2\\ & =9^2-{25}^2\\ & =81-625\\ & =-544\\ xy& =\boxed{{\displaystyle \boxed{{\displaystyle -138}}}}\end{array}$

Jadi, xy = −138.

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