Question

The number of prime numbers among the numbers $105!+2,105!+3,105!+4,...105!+104,105!+105$ is

• A. $31$
• B. $32$
• C. $33$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Finding the number of prime numbers among the numbers $\mathbf{105}\mathbf{!}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{,}\mathbf{}\mathbf{105}\mathbf{!}\mathbf{}\mathbf{+}\mathbf{}\mathbf{3}\mathbf{,}\mathbf{}\mathbf{105}\mathbf{!}\mathbf{}\mathbf{+}\mathbf{}\mathbf{4}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{105}\mathbf{!}\mathbf{}\mathbf{+}\mathbf{}\mathbf{104}\mathbf{,}\mathbf{}\mathbf{105}\mathbf{!}\mathbf{}\mathbf{+}\mathbf{}\mathbf{105}\mathbf{}$is

Given: $105!+2,105!+3,105!+4,...105!+104\text{,}105!+105$

Any set of numbers is divisible by number added to the factorial. Therefore, there is no prime numbers here.

$$\begin{gathered} 105!+2=2\times 3\times 4\times ...\times 105+2 \\ =2\times (1+3\times 4\times ...\times 105) \end{gathered}$$

The above number is even, it is not prime.

$$\begin{gathered} 105!+101=2\times 3\times ...\times 101\times ...\times 105+101 \\ =(2\times 3\times ...100\times 102\times 103\times 104\times 105+1)\times 101 \end{gathered}$$

Which is divisible by $101$ and hence not a prime number

Similarly the other number too are composite and therefore not prime.

Hence the correct answer is option (D).