Question

If ${{D}_{30}}$ is the set of all divisors of \[30\], \[x,y\in {{D}_{30}}\], we define $x+y=\text{LCM}\left( x,y \right)$, $x\cdot y=\text{GCD}\left( x,y \right)$, $x'=\dfrac{30}{x}$ and $f\left( x,y,z \right)=\left( x+y \right)\cdot \left( y'+z \right)$, then $f\left( 2,5,15 \right)$ is equal to
(a) $2$
(b) $5$
(c) \[10\]
(d) \[15\]

Answer 1

Hint: In the given question, three operators are defined for the numbers $x,y$ belonging to the set ${{D}_{30}}$, which includes all the divisors of \[30\]. Two of these are the binary operators $+$ and $\cdot $, and the third is the unary operator $'$. The function is given as $f\left( x,y,z \right)=\left( x+y \right)\cdot \left( y'+z \right)$, in which all of the three operators are present. For calculating the value of $f\left( 2,5,15 \right)$, we need to check whether the numbers $2,5,15$ belong to the set ${{D}_{30}}$ or not. If they do, then we will substitute $x=2,y=5,z=15$ in the function and solve the obtained expression using the definitions of the operators given in the question.

Complete step by step solution:
We know that the divisors of \[30\] are $1,2,3,5,6,10,15,30$. Therefore the set ${{D}_{30}}$ is given by
$\Rightarrow {{D}_{30}}\in \left\{ 1,2,3,5,6,10,15,30 \right\}........\left( i \right)$
Now, according to the question, we need to calculate the value of $f\left( 2,5,15 \right)$, where the function $f$ is defined as
\[\Rightarrow f\left( x,y,z \right)=\left( x+y \right)\cdot \left( y'+z \right).......(ii)\]
We can see in the definition for the function that it contains all of the three operators $+$, $\cdot $, and $'$ which are defined for the set ${{D}_{30}}$. Therefore before substituting $x=2,y=5,z=15$ into the given function, we need to check whether they belong to the set ${{D}_{30}}$ or not. We can clearly see from (i) that $2,5,15$ all belong to the set ${{D}_{30}}$. So the definition of all of the three operators will be applicable to them.
Therefore substituting $x=2,y=5,z=15$ in (i) we get
$\Rightarrow f\left( 2,5,15 \right)=\left( 2+5 \right)\cdot \left( 5'+15 \right)$
The operator $'$ is defined as $x'=\dfrac{30}{x}$. So we can write the above equation as
$\begin{align}
  & \Rightarrow f\left( 2,5,15 \right)=\left( 2+5 \right)\cdot \left( \dfrac{30}{5}+15 \right) \\
 & \Rightarrow f\left( 2,5,15 \right)=\left( 2+5 \right)\cdot \left( 6+15 \right) \\
\end{align}$
Now, the operator $+$ is defined as $x+y=\text{LCM}\left( x,y \right)$. So we can write the above equation as
\[\Rightarrow f\left( 2,5,15 \right)=\text{LCM}\left( 2,5 \right)\cdot \text{LCM}\left( 6,15 \right)\]
We know that the LCM of $2,5$ is $10$ and that of $6,15$ is $30$. Therefore we substitute $\text{LCM}\left( 2,5 \right)=10$ and $\text{LCM}\left( 6,15 \right)=30$ in the above equation to get
\[\Rightarrow f\left( 2,5,15 \right)=10\cdot 30\]
Finally, the operator \[\cdot \] is defined as $x\cdot y=\text{GCD}\left( x,y \right)$. So the above equation becomes
\[\Rightarrow f\left( 2,5,15 \right)=\text{GCD}\left( 10,30 \right)\]
The GCD or the HCF of $10,30$ is $10$, since $10$ is the factor of $30$. So we finally obtain
\[\Rightarrow f\left( 2,5,15 \right)=10\]

So, the correct answer is “Option c”.

Note: Do not consider the operators $+$ and $\cdot $ to be the addition and the multiplication operators. Since we are in a common practice to treat them, so we may be tempted to write $2+5=7$. The symbols for the operators given are intentionally of the mathematical operators of addition and multiplication. Also, do not forget to check whether the numbers, which are the argument to the function, belong to the given set. This is because then only the operator definition given in the question will be applicable to them. If they did not belong to the given set, then the usual definition of the respective operators would be applicable.