Question

$V$is a product of the first $41$ natural numbers $A=V+1$. The number of primes among $A+1,A+2,A+3,A+4,A+39,A+40$.

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is D

$0$

Step 1: Simplify the expression:

Given: $V$ is a product of the first $41$ natural numbers and $A=V+1$

$$\begin{gathered} \therefore A+1,A+2,A+3,A+4,A+39,A+40\mathbf{}\mathbf{[}\mathbf{\therefore }\mathbf{}\mathbf{G}\mathbf{i}\mathbf{v}\mathbf{e}\mathbf{n}\mathbf{:}\mathbf{}\mathbf{A}\mathbf{=}\mathbf{}\mathit{V}\mathbf{+}\mathbf{1}\mathbf{}\mathbf{]} \\ =V+1+1,V+1+2,V+1+3,V+1+4,V+1+39,V+1+40 \\ =V+2,V+3,V+4,V+5,V+40,V+41.........\left(1\right) \end{gathered}$$

Step 2: Solving the equation:

$$\begin{gathered} LetV+2,V+3,V+4,V+5,V+40,V+41\left[whereKisoneofthefirst41naturalnumbersbutnot1\right] \\ =V+K\left[K=2,3,4,............41\right] \\ Now, \\ \therefore Vistheproductoffirst41naturalnumbers\left[Given\right] \\ So,Visdivisblebyeachofthefollowingnaturalnumbers. \\ Thus,V+KisalsodivisblebyanyK\text{'}s. \\ eg:-V+2divisibleby2 \\ V+3divisibleby3 \\ V+41isdivisbleby41 \\ \because PrimenumberisdivisblebyitselfonlybuthereV+Kcontradicts. \\ so,V+Kisnotprimenumber \\ Therefore,\mathit{T}\mathit{h}\mathit{e}\mathbf{}\mathit{n}\mathit{u}\mathit{m}\mathit{b}\mathit{e}\mathit{r}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{p}\mathit{r}\mathit{i}\mathit{m}\mathit{e}\mathit{s}\mathbf{}\mathit{i}\mathit{s}\mathbf{}\mathbf{0}\mathbf{.} \end{gathered}$$

Hence, the final answer is an option D.