Question

Let $P_1$ be the set of all prime numbers, i.e., $P_1=(2,3,5,7,\mathrm{11......}),$ Let $P_n=\{np|p\in P_1\}$ i.e, the set of all prime multiples of $n$. Then which of the following sets in non empty?

• A. $P_1\cap P_{23}$
• B. $P_7\cap P_{21}$
• C. $P_{12}\cap P_{20}$
• D. $P_{20}\cap P_{24}$

Answer 1

The correct option is C $P_{12}\cap P_{20}$

It is given that $P_1=(2,3,5,7,\mathrm{11......})$.

Let us consider option $A$ that is

$P_1=(2,3,5,7,\mathrm{11......})$

$P_{23}=(46,69,115,161,......)$

$P_1\cap P_{23}$ has no common element.

Now consider option $B$

$P_7=(14,21,35,49,......)$

$P_{21}=(42,63,105,147,......)$

$P_7\cap P_{21}$ has no common element.

Now consider option $C$

$P_{12}=(24,36,60,84,......)$

$P_{20}=(40,60,100,140,......)$

$P_{12}\cap P_{20}$ has common element that is $60$.

Therefore $P_{12}\cap P_{20}$ is non empty.

Hence option $C$ is correct.