Question

Let $X$ and $Y$ be the sets of all positive divisors of $400$ and $1000$ respectively (including $1$ and the number). Then $n(X\cap Y)=$

• A. $4$
• B. $6$
• C. $8$
• D. $12$

Answer 1

The correct option is D

$12$

Explanation for the correct option:

Determine the value of $n(X\cap Y)$

The factors of $400=\left\{1,2,4,5,8,10,16,20,25,40,50,80,100,200,400\right\}$

The factors of $1000=\left\{1,2,4,5,8,10,20,25,40,50,100,125,200,250,500,1000\right\}$

Let $X$ be the sets of all positive divisors of $400$

$X=\left\{1,2,4,5,8,10,16,20,25,40,50,80,100,200,400\right\}$

Let $Y$ be the sets of all positive divisors of $1000$

$Y=\left\{1,2,4,5,8,10,20,25,40,50,100,125,200,250,500,1000\right\}$

Calculating $n(X\cap Y)$

$X\cap Y=\left\{x/x\in Xandx\in Y\right\}$

$X\cap Y=\left\{1,2,4,5,8,10,20,25,40,50,100,200\right\}$

$n(X\cap Y)=12$

Hence, option (D) is the correct answer