Question

# If the H.C.F of (x, y) =1, then the H.C.F of (x-y, x+y) =A. 1 or 2B. x or yC. x+y or x-yD. 4

Hint: Suppose any two random numbers whose H.C.F is 1, as you know H.C.F of two numbers is 1 if these numbers have no common factors except 1.

Let x=7 and y=5 as 5 and 7 have no common factors except 1 so H.C.F of these numbers is 1. The value of $x + y = 7 + 5 = 12$ and $x - y = 7 - 5 = 2$.
So, H.C.F of (x-y, x+y) =H.C.F of (2, 12). The factors of 2 are 2 and 1 and the factors of 12 is $1 \times 2 \times 2 \times 3$.
So the common factor of (2, 12) is $1 \times 2 = 2$.
Therefore H.C.F of (x-y, x+y) =2.
Now let another two numbers x=2 and y=1, as 2 and 1 have no common factors except 1 so H.C.F of these numbers is 1.
$\Rightarrow x + y = 2 + 1 = 3 \\ \Rightarrow x - y = 2 - 1 = 1 \\$
So H.C.F of (x-y, x+y) = H.C.F of (1, 3)
So the factors of 1 and the factors of 3 is $1 \times 3$. So the common factors of (1, 3) is 1.
So, H.C.F of (x-y, x+y) =1.
Therefore, H.C.F of (x-y, x+y) is 1 or 2.
So option â€˜aâ€™ is correct.

Note: In such types of questions always put values in place of x and y so that H.C.F of x and y is 1, then calculate the values of $x - y,x + y$ and then calculate the H.C.F of these numbers, then we will get the required answer.