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Consider numbers greater than one that satisfy the following properties:

  1. They have no repeated prime factors;
  2. For all primes $p \geq 2$, $p$ divides the number if and only if $p − 1$ divides the number.

The number of such numbers is

  1. $0$
  2. $5$
  3. $100$
  4. Infinite
  5. None of the above
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The prime factors of $30$ are $2,3,5$, so it satisfies the 1st constraint.

However, $\matrix{
(2-1) & \rm divides & 30 & \color{green}\checkmark\\
(3-1) & \rm divides & 30 & \color{green}\checkmark\\
(5-1) & \rm divides & 30 & \color{red}\times\\
}$ and thus it doesn't satisfy the 2nd constraint.
One can prove that for $n$ to satisfy these properties, $n=p(p−1)$ for some prime $p$ and that $(p−1)$ satisfies the properties too.

Some examples of the numbers that satisfy both constraints are:
$$\begin{align}
&2\\
2 \times 3 = &6\\
6 \times 7 = &42\\
42 \times 43 = &1806
\end{align}$$Now $1807$ is not a prime and hence breaks the sequence. So, number of such numbers is $4.$

Correct Answer: E.
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(a) no repeated prime number.So it can be{2,3,5,6(2*3 so no repeatation of prime in 6),7,11, 13,15,..........,}

(b) it can be (2,3)=6

                    (6,7)=42

                    ...............

                   (10,11)=110

So, we can say, as prime no is infinite, answer will be (d)infinite
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