Question

If $D_{30}$ is the set of the divisors of $30$, $x,y\in D_{30}$, we define $x+y=LCM(x,y),x·y=GCD(x,y),x\text{'}=\frac{30}{x}$ and $f(x,y,z)=(x+y)\times (y\text{'}+z)$, then $f(2,5,15)$ is equal to

• A. $2$
• B. $5$
• C. $10$
• D. $15$

Answer 1

The correct option is C

$10$

Find the value of $\mathit{f}\mathbf{}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{5}\mathbf{,}\mathbf{}\mathbf{15}\mathbf{)}$:

Given, $D_{30}$ is the set of the divisors of $30$

$\Rightarrow$$D_{30}=\left\{1,2,3,5,6,10,15,30\right\}$

Also we have,

$f(x,y,z)=(x+y)\times (y\text{'}+z)$

$\begin{array}{rcl}\therefore f(2,5,15)& =& (2+5)·(5’+15)\\ & =& (2+5)·\left(\frac{30}{5}+15\right)\\ & =& (2+5)·\left(6+15\right)\end{array}$ $\left[\because x\text{'}=\frac{30}{x}\right]$

$x+y=LCM(x,y)$

$\Rightarrow$$f(2,5,15)=LCM(2,5)·LCM(6,15)$

$=10·30$

$x·y=GCD(x,y)$

$\begin{array}{rcl}\therefore f(2,5,15)& =& GCD\left(10,30\right)\\ & =& \mathbf{10}\end{array}$

Hence, Option ‘C’ is Correct.