Question

V is the product of first natural number $ A=V+1 $ , the numbers of prime along
 $ A+1,\,A+2,\,A+3............A+40 $ is
 $ \begin{align}
  & a)1 \\
 & b)2 \\
 & c)3 \\
 & d)0 \\
\end{align} $

Answer 1

Hint: We are given that V is a product of the first 41 natural numbers then we have $ A=V+1 $, we use A as $ V+1 $ to rewrite $ A+1,\, A+2,\, A+3............A+40 $ as \[V+1,\, V+2,\, V+3............V+41\], we then use that since V is a product of first $ 41 $ natural numbers so it is divisible by all first $ 41 $ natural numbers. We will see that our terms are V+k type. Each term is divisible by k that helps in finding our answer.

Complete step by step answer:
We are given that V is denoted as the product of the first $ 41 $ natural numbers
Natural numbers are the terms that start from $ 1 $ and move on the positive side first $ 41 $ natural numbers are $ 1,2,3,4,...................41 $
Now we have V, V is the product of these natural Numbers.
So,
 $ V=1\times 2\times 3\times 4\times 5...........\times 39\times 40 $
So, we can say that V is divisible by each of the first $ 41 $ natural numbers.
Now we have
 $ A=V+1 $
Now we are asked to find the number of prime among $ A+1,\, A+2,\, A+3............A+40 $ first we use $ A=V+1 $ in each of the above terms. We will get they become \[V+1,\, V+2,\, V+3............V+40\]
Simplifying, we get
\[V+1,\,V+2,\,V+3............V+41\]
So we get that each term we had above is of the type V+k, where k is one of the first $ 41 $ natural numbers except $ 1 $, k is never $ 1 $.
Now as we know that as V is a product of the first $ 41 $ natural numbers so V is divisible by each of the natural numbers.
So our numbers are of type V+k, where V is divisible by any of the following k’s.
So, each V+k is divisible by k that $ V+2 $ is divisible by $ 2 $
 $ V+3 $ is divisible by $ 3 $
 $ V+4 $ is divisible by $ 4 $
And so, one till $ v+41 $ , which is divisible by $ 41 $ .
The definition of prime says a number that is not divisible by any number other than itself but here each term V+k is divisible by k.
So, none of the terms is a prime number.
So, number of prime among $ A+1,\,A+2,\,A+3............A+40 $ is $ 0 $
So the correct option is d) $ 0 $

Note:
 Remember that if x and y are both divisible by a number say z then their sum is also divisible by z. for example $ 4 $ and $ 8 $ are divisible by $ 2 $. So, $ 4+8=12 $ is also divisible by $ 2 $ . We use this to say that V+k is divisible by k, as V and k both are divisible by k.