Question:medium

Let \( p_1 \) and \( p_2 \) denote two arbitrary prime numbers. Which one of the following statements is correct for all values of \( p_1 \) and \( p_2 \)?

Show Hint

Product of two primes is always composite because it has at least three divisors: 1, each prime, and their product.
Updated On: Feb 9, 2026
  • \( p_1 + p_2 \) is not a prime number.
  • \( p_1 p_2 \) is not a prime number.
  • \( p_1 + p_2 + 1 \) is a prime number.
  • \( p_1 p_2 + 1 \) is a prime number.
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Clarify what is being asked. 
Two numbers $p_1$ and $p_2$ are given, and both are prime.
We must identify the statement that remains true for every possible choice of these prime numbers.

Step 2: Recall a basic fact about primes.
A prime number has exactly two positive divisors: 1 and itself.
Any number that has more than two divisors is therefore not prime.

Step 3: Test each option one by one.
 

(A) $p_1 + p_2$ is not a prime number.
This does not always hold.
For example, if $p_1=2$ and $p_2=3$, then:

\[ p_1+p_2 = 2+3 = 5 \]

Since 5 is prime, this statement fails in some cases.

(B) $p_1 p_2$ is not a prime number.
The product of two prime numbers is always composite.
It is divisible by $1$, $p_1$, $p_2$, and $p_1p_2$ itself.
Hence, it can never be prime, regardless of which primes are chosen.

(C) $p_1 + p_2 + 1$ is a prime number.
This is also not guaranteed.
Taking $p_1=2$ and $p_2=3$ gives:

\[ 2+3+1 = 6 \]

which is clearly not prime.

(D) $p_1 p_2 + 1$ is a prime number.
This fails as well.
For instance, with $p_1=3$ and $p_2=5$:

\[ 3\times5+1 = 16 \]

which is composite.

Step 4: Final conclusion.
The only statement that holds true for all choices of prime numbers is:

\[ \boxed{p_1p_2 \text{ is not a prime number}} \]

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