Binary Operation : Problems and Solutions

1.        State whether the operation $ \displaystyle x\odot y$ = the greatest integer less than $ \displaystyle \frac{x}{y}+ \frac{y}{x}$ on the set of positive integers is a binary operation. If it is a binary operation, find $ \displaystyle (3\odot2)$ and $ \displaystyle 2 \odot (3\odot4)$.

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        $ \displaystyle \ \ \ \ \ x\odot y=\text{the greatest integer less than}\ \frac{x}{y}+\frac{y}{x}$

        $ \displaystyle \begin{array}{l}\ \ \ \ \ \text{Since any image is the greatest integer,}\\\ \ \ \ \ \text{closure property is satisfied}\text{.}\\\\\therefore \ \ \ \odot \text{is a binary operation}\text{.}\end{array}$

        $ \displaystyle \ \ \ \ \ \text{3}\odot 2=\text{the greatest integer less than}\ \left( {\frac{3}{2}+\frac{2}{3}} \right)$

        $ \displaystyle \begin{array}{l}\therefore \ \ \ \text{3}\odot 2=\text{the greatest integer less than}\ \left( {2.167} \right)\\\\\therefore \ \ \ \text{3}\odot 2=2\end{array}$

        $ \displaystyle \ \ \ \ \ \text{3}\odot 4=\text{the greatest integer less than}\ \left( {\frac{3}{4}+\frac{4}{3}} \right)$

        $ \displaystyle \begin{array}{l}\therefore \ \ \ \text{3}\odot 4=\text{the greatest integer less than}\ \left( {2.083} \right)\\\\\therefore \ \ \ \text{3}\odot 4=2\ \ \end{array}$

        $ \displaystyle \ \ \ \ \ 2\odot \left( {\text{3}\odot 4} \right)=2\odot 2=\ \text{the greatest integer less than}\ \left( {\frac{2}{2}+\frac{2}{2}} \right)$

        $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ \text{the greatest integer less than}\ 2\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\ 1\ \ \ \ \end{array}$


2.        Let $ \displaystyle g:{{Z}^{+}}\times {{Z}^{+}}\to N$ be defined by $ \displaystyle (x, y)\mapsto g (x , y)$ = the remainder when $ \displaystyle x^y$ is divided by $ \displaystyle 3$. Find $ \displaystyle g(g(2, 3),\ 4)$. ($ \displaystyle {Z}^{+}$ = the set of positive integers and $ \displaystyle N$ = the set of natural numbers)

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        $ \displaystyle \begin{array}{l}\ \ \ \ g:{{Z}^{+}}\times {{Z}^{+}}\to N\\\\\ \ \ \ g(x,y)=\text{the remainder when }{{x}^{y}}\text{ is divided by 3}\\\\\therefore \ \ g(2,3)=\text{the remainder when }{{2}^{3}}\ \text{is divided by 3}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\\\\\therefore \ \ g(g(2,3),4)=g(\text{2},4)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\text{the remainder when }{{2}^{4}}\text{ is divided by 3}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =1\end{array}$


3.        Let $ \displaystyle f : R\to R$ be defined by $ \displaystyle f(x) = x^3$ and $ \displaystyle g : N\to N$ be defined by $ \displaystyle g(y) = y^2 +y$. Define $ \displaystyle x\odot y = f(x) g(y)$. $ \displaystyle \odot$ is to be a binary operation, find possible domain and codomain. If $ \displaystyle \odot$ is a binary operation, prove it. Determine that $ \displaystyle \odot$ iscommutative or not.

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        $ \displaystyle \begin{array}{l}\ \ \ \ f:R\to R,\ f(x)={{x}^{3}}\\\\\ \ \ \ g:R\to R,\ g(y)={{y}^{2}}+y\\\\\ \ \ \ x\odot y=f(x)g(y)\\\\\therefore \ \ x\odot y={{x}^{3}}\left( {{{y}^{2}}+y} \right)\\\\\ \ \ \ \text{Since}\ {{x}^{3}}\in R\ \text{for every }x\in R\ \text{and}\\\\\ \ \ \ {{y}^{2}}+y\in R\ \text{for every }y\in R,\\\\\therefore \ \ \text{Domain of }x\odot y=R\times R\\\\\ \ \ \ \text{Codomain of }x\odot y=R\\\\\therefore \ \ \odot :R\times R\to R\\\\\therefore \ \ \text{Closure property is satisfied}\text{.}\\\\\therefore \ \ \odot \text{ is a binary operation}\text{.}\\\\\ \ \ \ 1\odot 2={{1}^{3}}\left( {{{2}^{2}}+2} \right)=6\\\\\ \ \ \ 2\odot 1={{2}^{3}}\left( {{{1}^{2}}+1} \right)=16\\\\\therefore \ \ \odot \text{ is not commutative}\text{.}\end{array}$


4.        A binary operation $ \displaystyle \odot$ on the set $ \displaystyle R$ of real numbers is defined by $ \displaystyle a\odot b = a + b + ab$. Show that the binary operation $ \displaystyle \odot$ is (i) commutative (ii) associative.

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        $ \displaystyle \begin{array}{l}\ \ \ \ a\odot b=a+b+ab\ \text{where}\ a\in R,b\in R\\\\\ \ \ \ b\odot a=b+a+ba=a+b+ab\\\\\therefore \ \ a\odot b=b\odot a\\\\\therefore \ \ \odot \text{ is commutative}\text{.}\\\\\therefore \ \ \ \left( {a\odot b} \right)\odot c=(a+b+ab)\odot c\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =(a+b+ab)+c+(a+b+ab)c\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b+ab+c+ac+bc+abc\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b+c+ab+ac+bc+abc\\\\\therefore \ \ \ a\odot \left( {b\odot c} \right)=a\odot (b+c+bc)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b+c+bc+a(b+c+bc)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b+c+bc+ab+ac+abc\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =a+b+c+ab+ac+bc+abc\\\\\therefore \ \ \ \left( {a\odot b} \right)\odot c=\ a\odot \left( {b\odot c} \right)\\\\\therefore \ \ \odot \text{ is associative}\text{.}\end{array}$


5.        Let $ \displaystyle Z$ be the set of all integers and $ \displaystyle A = \left\{ { 0 , 1 } \right\}.$ Let $ \displaystyle {{\odot }_{1}}:A\times A\to Z$ be defined by $ \displaystyle(x , y )\mapsto \odot_1 (x , y) = x^2 + y$. Let $ \displaystyle {{\odot }_{2}}:A\times A\to Z$ be defined by $ \displaystyle (x , y )\mapsto \odot_2 (x , y) = x^2 y$. Is $ \displaystyle \odot_1$ a binary operation on $ \displaystyle A$? Why? Is $ \displaystyle \odot_2$ a binary operation on $ \displaystyle A$? Why ?

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        $ \displaystyle \begin{array}{l}\ \ \ \ A=\left\{ {0,1} \right\}\\\\\ \ \ \ {{\odot }_{1}}:A\times A\to Z,\\\\\ \ \ \ (x,y)\mapsto {{\odot }_{1}}(x,y)={{x}^{2}}+y\\\\\ \ \ \ {{\odot }_{2}}:A\times A\to Z\\\\\ \ \ \ (x,y)\mapsto {{\odot }_{2}}(x,y)={{x}^{2}}y\\\\\ \ \ \ A\times A=\left\{ {(0,0),(0,1),(1,0),(1,1)} \right\}\\\\\ \ \ \ \ x{{\odot }_{1}}\ y={{x}^{2}}+y\\\\\therefore \ \ \ 0{{\odot }_{1}}0={{0}^{2}}+0=0\\\\\ \ \ \ \ 0{{\odot }_{1}}1={{0}^{2}}+1=1\\\\\ \ \ \ \ 1{{\odot }_{1}}0={{1}^{2}}+0=1\\\\\ \ \ \ \ 1{{\odot }_{1}}1={{1}^{2}}+1=2\\\\\therefore \ \ \ \text{Range of }{{\odot }_{1}}=\{0,1,2\}\\\\\therefore \ \ \ \text{Range of }{{\odot }_{1}}\not\subset A\\\\\therefore \ \ \ \text{Closure property is not satisfied}\text{.}\\\\\therefore \ \ \ {{\odot }_{1}}\ \text{is not a binary operation}\text{.}\\\\\ \ \ \ \ x{{\odot }_{2}}\ y={{x}^{2}}y\\\\\therefore \ \ \ 0{{\odot }_{2}}0={{0}^{2}}(0)=0\\\\\ \ \ \ \ 0{{\odot }_{2}}1={{0}^{2}}(1)=0\\\\\ \ \ \ \ 1{{\odot }_{2}}0={{1}^{2}}(0)=0\\\\\ \ \ \ \ 1{{\odot }_{2}}1={{1}^{2}}(1)=1\\\\\therefore \ \ \ \text{Range of }v=\{0,1\}\\\\\therefore \ \ \ \text{Range of }{{\odot }_{2}}\subset A\\\\\therefore \ \ \ \text{Closure property is satisfied}\text{.}\\\\\therefore \ \ \ {{\odot }_{2}}\ \text{is a binary operation}\text{.}\end{array}$


6.        The operation $ \displaystyle \odot$ on $ \displaystyle R$ is defined by $ \displaystyle x\odot y=\frac{{{{x}^{2}}+{{y}^{2}}}}{2}-xy$ for all real numbers $ \displaystyle x$ and $ \displaystyle y$. Prove that $ \displaystyle \odot$ is commutative but not associative.

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     $ \displaystyle \ \ \ \ x\odot y=\frac{{{{x}^{2}}+{{y}^{2}}}}{2}-xy,\ x,y\in R$

     $ \displaystyle \ \ \ \ y\odot x=\frac{{{{y}^{2}}+{{x}^{2}}}}{2}-yx$

     $ \displaystyle \therefore \ \ y\odot x=\frac{{{{x}^{2}}+{{y}^{2}}}}{2}-xy$

     $ \displaystyle \begin{array}{l}\therefore \ \ x\odot y=y\odot x\\\\\therefore \ \ \odot \ \text{is commutative}.\end{array}$

     $ \displaystyle \ \ \ \ 1\odot 2=\frac{{{{1}^{2}}+{{2}^{2}}}}{2}-1(2)=\frac{1}{2}$

     $ \displaystyle \ \ \ \ \left( {1\odot 2} \right)\odot 3=\frac{1}{2}\odot 3$

     $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{\left(      {\frac{1}{2}} \right)}}^{2}}+{{3}^{2}}}}{2}-\frac{1}{2}(3)$

     $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{25}}{8}$

     $ \displaystyle \ \ \ \ 2\odot 3=\frac{{{{2}^{2}}+{{3}^{2}}}}{2}-2(3)=\frac{1}{2}$

     $ \displaystyle \ \ \ \ 1\odot \left( {2\odot 3} \right)=1\odot \frac{1}{2}$

     $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{1}^{2}}+{{{\left( {\frac{1}{2}} \right)}}^{2}}}}{2}-1\left( {\frac{1}{2}} \right)$

     $ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{8}$

     $ \displaystyle \begin{array}{l}\therefore \ \ \left( {1\odot 2} \right)\odot 3\ne 1\odot \left( {2\odot 3} \right)\\\\\therefore \ \ \odot \ \text{is not associative}.\end{array}$


7.        The operation $ \displaystyle \odot$ is defined by $ \displaystyle (2a + b) \odot(a + 2b) = a^2 + ab + b^2$. Find $ \displaystyle 6\odot 9$.

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     $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ (2a+b)\odot (a+2b)={{a}^{2}}+ab+{{b}^{2}}\\\\\ \ \ \ \ \ \ \ \text{Let}\ 2a+b=6\ \ \ \ \ \ \ \ \ \ \ \ -----(1)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ a+2b=9\ \ \ \ \ \ \ \ \ \ \ -----(2)\\\\\ \ \ \ \ \ \ \ \text{By Equation(1)+Equation(2),}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ 3a+3b=15\ \ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ a+b=5\ \ \ \ \ \ \ \ \ \ \ \ \ -----(3)\\\\\ \ \ \ \ \ \ \ \text{By Equation(1)}-\text{Equation(2),}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ a-b=-3\ \ \ \ \ \ \ \ \ \ \ -----(4)\\\\\ \ \ \ \ \ \ \ \text{By Equation(3)}+\text{Equation(4),}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ 2a=2\ \Rightarrow a=1\\\\\ \ \ \ \ \ \ \ \text{By Equation(3)}-\text{Equation(4),}\ \\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ 2b=8\ \Rightarrow b=4\\\\\ \ \ \ \ \ \ \ \therefore \ 6\odot 9=\left[ {2(1)+4} \right]\odot \left[ {1+2(4)} \right]\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{(1)}^{2}}+(1)(4)+{{(4)}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =21\end{array}$


8.        An operation $ \displaystyle \odot$ is defined by $ \displaystyle a\odot b = a^2 + ab + b^2,\ a,\ b\in R$. Solve the equation $ \displaystyle (6\odot k) - (k\odot 2) = 8 - 8k$.

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     $ \displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ a\odot b={{a}^{2}}+ab+{{b}^{2}}\\\\\ \ \ \ \ \ \ \ 6\odot k={{6}^{2}}+6k+{{k}^{2}}\\\\\ \ \ \ \ \ \ \ k\odot 2={{k}^{2}}+2k+{{2}^{2}}\\\\\ \ \ \ \therefore \ \ (6\odot k)-(k\odot 2)=8-8k\\\\\ \ \ \ \therefore \ \ ({{6}^{2}}+6k+{{k}^{2}})-({{k}^{2}}+2k+{{2}^{2}})=8-8k\\\\\ \ \ \ \therefore \ \ 4k-32=8-8k\\\\\ \ \ \ \therefore \ \ 12k=40\end{array}$

     $\displaystyle \ \ \ \ \therefore \ \ k=\frac{{10}}{3}$


9.        A binary operation $ \displaystyle \odot$ on $ \displaystyle R$ is defined by $ \displaystyle x\odot y = (2x - 3y)^2 - 5y^2$. Show that the binary operation is commutative. Find the values of $ \displaystyle k$ for which $ \displaystyle(-2)\odot k = 80$.

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     $\displaystyle \begin{array}{l}\ \ \ \ x\odot y={{(2x-3y)}^{2}}-5{{y}^{2}}\\\\\ \ \ \ x\odot y=4{{x}^{2}}-12xy+9{{y}^{2}}-5{{y}^{2}}\\\\\therefore \ \ x\odot y=4{{x}^{2}}-12xy+4{{y}^{2}}\\\\\therefore \ \ x\odot y=4\left( {{{x}^{2}}-3xy+{{y}^{2}}} \right)\\\\\therefore \ \ y\odot x=4\left( {{{y}^{2}}-3yx+{{x}^{2}}} \right)\\\\\therefore \ \ y\odot x=4\left( {{{x}^{2}}-3xy+{{y}^{2}}} \right)\\\\\therefore \ \ x\odot y=y\odot x\\\\\therefore \ \ \odot \ \text{is commutative}.\\\\\ \ \ \ (-2)\odot k=80\ \ \left[ {\text{given}} \right]\\\\\therefore \ \ 4\left[ {{{{(-2)}}^{2}}-3(-2)k+{{k}^{2}}} \right]=80\\\\\therefore \ \ {{k}^{2}}+6k-16=0\\\\\therefore \ \ (k+8)(k-2)=0\\\\\therefore \ \ k=-8\ \text{or}\ k=2\end{array}$


10.     Let $ \displaystyle A=\{0,\ 1,\ 2,\ 3,\ 4\}$. The binary operation $ \displaystyle {{\oplus }_{5}}$ on the set $ \displaystyle A$ is defined by $ \displaystyle x\ {{\oplus }_{5}}\ y$ = the remainder when $ \displaystyle x + 2y$ is divided by 5. Make a Cayley table.

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     $ \displaystyle \begin{array}{l}\ \ A=\{0,\ 1,\ 2,\ 3,\ 4\}\\\\\ \ x\ {{\oplus }_{5}}\ y=\text{the remainder when}\ x+2y\ \text{is divided by 5}\text{.}\\\\\ \ 0\ {{\oplus }_{5}}\ 0=0\ \ \ \ \ \ 1\ {{\oplus }_{5}}\ 0=1\ \ \ \ \ 2\ {{\oplus }_{5}}\ 0=2\ \ \ \ \ 3\ {{\oplus }_{5}}\ 0=3\ \ \ \ \ \ 4\ {{\oplus }_{5}}\ 0=4\ \\\\\ \ 0\ {{\oplus }_{5}}\ 1=2\ \ \ \ \ \ 1\ {{\oplus }_{5}}\ 1=3\ \ \ \ \ \ 2\ {{\oplus }_{5}}\ 1=4\ \ \ \ \ 3\ {{\oplus }_{5}}\ 1=0\ \ \ \ \ \ 4\ {{\oplus }_{5}}\ 1=1\ \\\\\ \ 0\ {{\oplus }_{5}}\ 2=4\ \ \ \ \ 1\ {{\oplus }_{5}}\ 2=0\ \ \ \ \ \ 2\ {{\oplus }_{5}}\ 2=1\ \ \ \ \ 3\ {{\oplus }_{5}}\ 2=2\ \ \ \ \ 4\ {{\oplus }_{5}}\ 2=3\ \\\\\ \ 0\ {{\oplus }_{5}}\ 3=1\ \ \ \ \ \ 1\ {{\oplus }_{5}}\ 3=2\ \ \ \ \ \ 2\ {{\oplus }_{5}}\ 3=3\ \ \ \ \ 3\ {{\oplus }_{5}}\ 3=4\ \ \ \ \ 4\ {{\oplus }_{5}}\ 3=0\ \\\\\ \ 0\ {{\oplus }_{5}}\ 4=3\ \ \ \ \ 1\ {{\oplus }_{5}}\ 4=4\ \ \ \ \ \ 2\ {{\oplus }_{5}}\ 4=0\ \ \ \ \ 3\ {{\oplus }_{5}}\ 4=1\ \ \ \ \ 4\ {{\oplus }_{5}}\ 4=2\ \end{array}$

     $ \displaystyle \text {Cayley Table}$
 ⊕5  0  1  2  3 4
 0  0 ⊕50  0 ⊕51 0 ⊕52 0 ⊕53  0 ⊕54
 1  1 ⊕50 1 ⊕51  1 ⊕52 1 ⊕53  1 ⊕54
 2  2 ⊕50 2 ⊕51 2 ⊕52 2 ⊕53 2 ⊕54
 3 3 ⊕50 3 ⊕51 3 ⊕52 3 ⊕53 3 ⊕54
 4  4 ⊕50 4 ⊕51 4 ⊕52 4 ⊕53 4 ⊕54


5      0      1      2     3     4
   0      0      2      4     1     3
   1      1      3      0     2     4
   2      2      4      1     3     0
   3      3      0      2     4     1
   4      4      1      3     0     2  



11.     The binary operation $ \displaystyle \odot_1$ and $ \displaystyle \odot_2$ on R defined by $ \displaystyle x\odot_1 y = x^2 - y^2$ and $ \displaystyle x\odot_2 y = 7x + 4y$. Find $ \displaystyle(2 \odot_2 1)\odot_1 4$. Find also $ \displaystyle x$ if $ \displaystyle (- 3\odot_1 2) \odot_2 (1\odot_1 x) = 3$.

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.     $ \displaystyle \begin{array}{l}\ \ \ \ x{{\odot }_{1}}y={{x}^{2}}-{{y}^{2}}\\\\\ \ \ \ x{{\odot }_{2}}y=7x+4y\\\\\therefore \ \ \ 2{{\odot }_{2}}1=7(2)+4(1)=18\\\\\therefore \ \ \ \left( {2{{\odot }_{2}}1} \right){{\odot }_{1}}4=18{{\odot }_{1}}4\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{18}^{2}}-{{4}^{2}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =306\\\\\ \ \ \ \ -3{{\odot }_{1}}2={{(-3)}^{2}}-{{2}^{2}}=5\\\\\ \ \ \ \ 1{{\odot }_{1}}x={{1}^{2}}-{{x}^{2}}=1-{{x}^{2}}\\\\\ \ \ \ \ \text{By the problem,}\\\\\ \ \ \ \ (-3{{\odot }_{1}}2){{\odot }_{2}}(1{{\odot }_{1}}x)=3\\\\\therefore \ \ \ 5{{\odot }_{2}}(1-{{x}^{2}})=3\\\\\ \ \ \ 7(5)+4(1-{{x}^{2}})=3\\\\\therefore \ \ \ 39-{{x}^{2}}=3\Rightarrow {{x}^{2}}=36\Rightarrow x=\pm 6\end{array}$

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