Analyzing Divisibility Statements involving Factorials

The question asks us to evaluate the correctness of two statements regarding the divisibility of expressions involving factorials.

We will analyze each statement separately, keeping in mind the concept of divisibility and related theorems like Wilson's Theorem.

Understanding Wilson's Theorem

Wilson's Theorem provides a criterion for primality. It states that a natural number $p > 1$ is a prime number if and only if $(p-1)! \equiv -1 \pmod{p}$.

This can also be written as $(p-1)! + 1$ is divisible by $p$, where $p$ is a prime number.

Analyzing Statement 1: (25)! + 1 is divisible by 26

Statement 1 concerns the divisibility of $(25)! + 1$ by 26.

• The number we are checking divisibility by is 26.
• First, let's check if 26 is a prime number. The factors of 26 are 1, 2, 13, and 26. Since 26 has factors other than 1 and itself, 26 is a composite number, not a prime number.
• Since 26 is not prime, Wilson's Theorem does not directly apply in the form $(p-1)! + 1$ being divisible by $p$.
• Let's consider $(25)! = 25 \times 24 \times \dots \times 1$.
• The number 26 is a composite number, $26 = 2 \times 13$.
• Since 25! contains the factors 2 and 13 within its product ($25! = \dots \times 13 \times \dots \times 2 \times \dots$), it must be divisible by $2 \times 13$, which is 26.
• So, $(25)! \equiv 0 \pmod{26}$.
• Now, let's consider $(25)! + 1$.
• $(25)! + 1 \equiv 0 + 1 \pmod{26}$
• $(25)! + 1 \equiv 1 \pmod{26}$
• This means that when $(25)! + 1$ is divided by 26, the remainder is 1.
• For $(25)! + 1$ to be divisible by 26, the remainder should be 0. Since the remainder is 1, $(25)! + 1$ is not divisible by 26.

Therefore, Statement 1 is incorrect.

Analyzing Statement 2: (6)! + 1 is divisible by 7

Statement 2 concerns the divisibility of $(6)! + 1$ by 7.

• The number we are checking divisibility by is 7.
• First, let's check if 7 is a prime number. The only factors of 7 are 1 and 7. Thus, 7 is a prime number.
• Since 7 is a prime number, we can apply Wilson's Theorem.
• According to Wilson's Theorem, for a prime $p$, $(p-1)! + 1$ is divisible by $p$.
• Here, $p = 7$. So, $(7-1)! + 1 = (6)! + 1$ should be divisible by 7.
• Let's verify this directly:
  • $(6)! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$
  • $(6)! + 1 = 720 + 1 = 721$
  • Now, let's divide 721 by 7: $721 \div 7$.
  • $721 = 700 + 21$
  • $700 \div 7 = 100$
  • $21 \div 7 = 3$
  • So, $721 \div 7 = 100 + 3 = 103$.
  • Since 721 divided by 7 gives a whole number quotient (103) with a remainder of 0, 721 is divisible by 7.
• Thus, $(6)! + 1$ is divisible by 7.

Therefore, Statement 2 is correct.

Conclusion on the Statements

Based on our analysis:

• Statement 1: (25)! + 1 is divisible by 26 (Incorrect)
• Statement 2: (6)! + 1 is divisible by 7 (Correct)

We are asked which of the above statements is/are correct. Only Statement 2 is correct.

Revision Table: Key Concepts for Divisibility and Factorials

Additional Information on Factorial Divisibility

While Wilson's Theorem is specific to prime numbers, the divisibility of $(n-1)! + 1$ by $n$ can be explored for composite numbers as well.

• If $n$ is a composite number greater than 4, then $(n-1)!$ is divisible by $n$.
  • Let $n = ab$ where $1 < a, b < n$.
  • If $a \ne b$, then both $a$ and $b$ are distinct factors in the product $(n-1)! = (n-1) \times \dots \times a \times \dots \times b \times \dots \times 1$. Thus, $(n-1)!$ is divisible by $ab = n$.
  • If $n = p^2$ for some prime $p$, and $n > 4$ (so $p > 2$), then $p < 2p < p^2$. Both $p$ and $2p$ are factors in $(n-1)! = (p^2 - 1)!$. Since $(p^2-1)!$ contains both $p$ and $2p$, it contains the factor $2p^2$, which is divisible by $p^2=n$. Thus, $(n-1)!$ is divisible by $n$. This holds for $n=9$ (3<6<9), $n=25$ (5<10<25), etc.
  • The exceptions for $(n-1)!$ being divisible by $n$ for composite $n$ are $n=4$. For $n=4$, $(4-1)! = 3! = 6$, which is not divisible by 4.
• Based on this, for composite $n > 4$, we have $(n-1)! \equiv 0 \pmod{n}$. Therefore, $(n-1)! + 1 \equiv 0 + 1 \equiv 1 \pmod{n}$. This means $(n-1)! + 1$ is NOT divisible by $n$ for composite $n > 4$.
• For $n=4$, $(4-1)! + 1 = 3! + 1 = 6 + 1 = 7$. 7 is not divisible by 4. So it also holds for $n=4$.
• In summary, $(n-1)! + 1$ is divisible by $n$ IF AND ONLY IF $n$ is a prime number. This is the converse of Wilson's Theorem and completes the statement of the theorem.
• In Statement 1, $n=26$, which is composite and $26 > 4$. So $(25)!$ is divisible by 26, and $(25)! + 1$ is not divisible by 26.
• In Statement 2, $n=7$, which is prime. So $(6)! + 1$ is divisible by 7, by Wilson's Theorem.