Question

If $D_{30}$ is the set of all divisors of $30,x,yϵD_{30}$, we define $x+y=LCM(x,y),x\cdot y=GCD(x,y),x^{\prime }=\frac{30}{x}$ and $f(x,y,z)=(x+y)\cdot (y^{\prime }+z)$, then $f(2,5,15)$ is equal to

• A. $2$
• B. $5$
• C. $10$
• D. $15$

Answer 1

Answer: (C)

$10$

The divisors of 30 are, $D_{30}=1,2,3,5,6,10,15,30$.

$f\left(2,5,15\right)=\left(2+5\right)\cdot \left(5^{\prime }+15\right)$

$=\left(10\right)\cdot \left(\frac{30}{5}+15\right)$

$=\left(10\right)\cdot \left(30\right)$

$=10$