If D30 is the set of all divisors of 30, x, y ∈ D30 , we define x + y = LCM(x, y), x ⋅ y = GCD (x, y) , x' = (30/x) and f(x, y, z) = (x + y) ⋅ (y' + z) , then f(2, 5, 15) is equal to

Question

If D30 is the set of all divisors of 30, x, y ∈ D30 , we define x + y = LCM(x, y), x ⋅ y = GCD (x, y) , x' = (30/x) and f(x, y, z) = (x + y) ⋅ (y' + z) , then f(2, 5, 15) is equal to (A)  2  (B)  5  (C)  10  (D)  15

Tardigrade Admin · Accepted Answer

D30= 1,2,3, 5,6,10,15,30 f(2,5,15) =(2+5) ⋅(5'+15) = 10 ⋅((30/5)+15) = 10(6+15) = 10 ⋅ 30 = 10

Q. If $D_{30}$ is the set of all divisors of $30, x, y \in D_{30}$ , we define $x + y = L CM (x, y), x \cdot y = GC D (x, y)$ , $x^{'} = \frac{30}{x}$ and $f (x, y, z) = (x + y) \cdot (y^{'} + z)$ , then $f (2, 5, 15)$ is equal to

$2$

$5$

$10$

$15$

$D_{30} = {1, 2, 3, 5, 6, 10, 15, 30}$
$f (2, 5, 15) = (2 + 5) \cdot (5^{'} + 15)$
$= 10 \cdot (\frac{30}{5} + 15)$
$= 10 (6 + 15)$
$= 10 \cdot 30$
$= 10$

Q. If D30​ is the set of all divisors of 30,x,y∈D30​ , we define x+y=LCM(x,y),x⋅y=GCD(x,y) , x′=x30​ and f(x,y,z)=(x+y)⋅(y′+z) , then f(2,5,15) is equal to

2

5

10

15

D30​={1,2,3,5,6,10,15,30} f(2,5,15)=(2+5)⋅(5′+15) =10⋅(530​+15) =10(6+15) =10⋅30 =10

Q. If $D_{30}$ is the set of all divisors of $30, x, y \in D_{30}$ , we define $x + y = L CM (x, y), x \cdot y = GC D (x, y)$ , $x^{'} = \frac{30}{x}$ and $f (x, y, z) = (x + y) \cdot (y^{'} + z)$ , then $f (2, 5, 15)$ is equal to

$2$

$5$

$10$

$15$

$D_{30} = {1, 2, 3, 5, 6, 10, 15, 30}$
$f (2, 5, 15) = (2 + 5) \cdot (5^{'} + 15)$
$= 10 \cdot (\frac{30}{5} + 15)$
$= 10 (6 + 15)$
$= 10 \cdot 30$
$= 10$