Consider the following statements: 1. \(\frac{n!}{3!}\) is divisible by 6, where n > 3 2. \(\frac{n!}{3!}+3 \) is divisible by 7, where n > 3 Which of the above statements is/are correct?
Neither 1 nor 2
Analyzing Divisibility of Factorial Expressions
This question asks us to evaluate the correctness of two statements regarding the divisibility of expressions involving factorials for values of n greater than 3.
Let's break down each statement carefully.
Statement 1 Analysis: Divisibility of \( \frac{n!}{3!} \) by 6 for n > 3
The expression is \( \frac{n!}{3!} \). For n > 3, we can write out the factorial expansion:
\( \frac{n!}{3!} = \frac{n \times (n-1) \times (n-2) \times \dots \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} \)
Simplifying, we get:
\( \frac{n!}{3!} = n \times (n-1) \times (n-2) \times \dots \times 4 \)
This is the product of integers from 4 up to n. The statement claims this product is divisible by 6 for all n > 3.
Let's test some values of n > 3:
- For n = 4: \( \frac{4!}{3!} = 4 \). Is 4 divisible by 6? No. \( 4 \div 6 \) leaves a remainder.
- For n = 5: \( \frac{5!}{3!} = 5 \times 4 = 20 \). Is 20 divisible by 6? No. \( 20 = 3 \times 6 + 2 \).
- For n = 6: \( \frac{6!}{3!} = 6 \times 5 \times 4 = 120 \). Is 120 divisible by 6? Yes. \( 120 = 20 \times 6 \).
For an integer to be divisible by 6, it must be divisible by both 2 and 3.
Consider the expression \( n \times (n-1) \times \dots \times 4 \). This is a product of consecutive integers (starting from 4 up to n) if n is large enough, or just the number 4 itself if n=4, or \(5 \times 4\) if n=5.
- If n = 4, the value is 4, which is divisible by 2 but not 3. So, not divisible by 6.
- If n = 5, the value is 20, which is divisible by 2 but not 3. So, not divisible by 6.
- If n = 6, the value is \(6 \times 5 \times 4 = 120\). This product contains 6 as a factor, so it is divisible by 6.
- If n > 6, the product \( n \times (n-1) \times \dots \times 4 \) includes the factor 6 (since it's \(n \times \dots \times 6 \times \dots \times 4\)). Any number with 6 as a factor is divisible by 6.
The statement claims divisibility by 6 for all n > 3. Since we found instances (n=4, n=5) where \( \frac{n!}{3!} \) is not divisible by 6, Statement 1 is false.
Statement 2 Analysis: Divisibility of \( \frac{n!}{3!}+3 \) by 7 for n > 3
The expression is \( \frac{n!}{3!}+3 \). The statement claims this expression is divisible by 7 for all n > 3.
Again, let's test some values of n > 3:
- For n = 4: \( \frac{4!}{3!}+3 = 4+3 = 7 \). Is 7 divisible by 7? Yes.
- For n = 5: \( \frac{5!}{3!}+3 = (5 \times 4)+3 = 20+3 = 23 \). Is 23 divisible by 7? No. \( 23 = 3 \times 7 + 2 \).
- For n = 6: \( \frac{6!}{3!}+3 = (6 \times 5 \times 4)+3 = 120+3 = 123 \). Is 123 divisible by 7? No. \( 123 = 17 \times 7 + 4 \).
Let's consider the behavior of \( \frac{n!}{3!} \) for larger n values with respect to divisibility by 7.
Recall \( \frac{n!}{3!} = n \times (n-1) \times \dots \times 4 \).
- If n = 4, \( \frac{4!}{3!} = 4 \). \( \frac{4!}{3!}+3 = 4+3 = 7 \), which is divisible by 7.
- If n = 5, \( \frac{5!}{3!} = 5 \times 4 = 20 \). \( \frac{5!}{3!}+3 = 20+3 = 23 \), not divisible by 7.
- If n = 6, \( \frac{6!}{3!} = 6 \times 5 \times 4 = 120 \). \( \frac{6!}{3!}+3 = 120+3 = 123 \), not divisible by 7.
- If n = 7, \( \frac{7!}{3!} = 7 \times 6 \times 5 \times 4 \). This product includes 7 as a factor, so \( \frac{7!}{3!} \) is divisible by 7. Thus, \( \frac{7!}{3!} \equiv 0 \pmod{7} \). Then \( \frac{7!}{3!}+3 \equiv 0+3 \equiv 3 \pmod{7} \). This is not divisible by 7.
- If n > 7, the product \( \frac{n!}{3!} = n \times (n-1) \times \dots \times 7 \times \dots \times 4 \) includes the factor 7. Therefore, for n \(\ge\) 7, \( \frac{n!}{3!} \) is divisible by 7, meaning \( \frac{n!}{3!} \equiv 0 \pmod{7} \). Consequently, for n \(\ge\) 7, \( \frac{n!}{3!}+3 \equiv 0+3 \equiv 3 \pmod{7} \). This is not divisible by 7.
The statement claims divisibility by 7 for all n > 3. While it is true for n=4, it is false for n=5, n=6, and all n \(\ge\) 7. Since it is not true for all n > 3, Statement 2 is false.
Conclusion
Based on our analysis, both Statement 1 and Statement 2 are false because we found values of n > 3 for which each statement does not hold true.
Therefore, neither of the given statements is correct.
| n | \( \frac{n!}{3!} \) | Divisible by 6? (Statement 1) | \( \frac{n!}{3!}+3 \) | Divisible by 7? (Statement 2) |
|---|---|---|---|---|
| 4 | 4 | No | 7 | Yes |
| 5 | 20 | No | 23 | No |
| 6 | 120 | Yes | 123 | No |
| 7 | \( 7 \times 6 \times 5 \times 4 = 840 \) | Yes | \( 840 + 3 = 843 \) | No ( \( 843 = 120 \times 7 + 3 \) ) |
Summary of Findings
- Statement 1 is false because \( \frac{n!}{3!} \) is not divisible by 6 for n=4 and n=5.
- Statement 2 is false because \( \frac{n!}{3!}+3 \) is not divisible by 7 for n=5, n=6, and all n \(\ge\) 7.
Thus, neither statement is correct.
Revision Table: Factorial Divisibility Concepts
| Concept | Description | Example |
|---|---|---|
| Factorial (n!) | The product of all positive integers up to n. \( n! = n \times (n-1) \times \dots \times 1 \). | \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \) |
| Simplifying Factorial Ratios | \( \frac{n!}{k!} = n \times (n-1) \times \dots \times (k+1) \) for \( n > k \). | \( \frac{7!}{4!} = 7 \times 6 \times 5 \) |
| Divisibility Rules | Rules to check if a number is exactly divisible by another number without remainder. | A number is divisible by 6 if it's divisible by both 2 and 3. |
| Modular Arithmetic | A system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value—the modulus. \( a \equiv b \pmod{m} \) means \( a-b \) is divisible by \( m \). | \( 23 \equiv 2 \pmod{7} \) because \( 23 - 2 = 21 \), which is divisible by 7. |
Additional Information: Properties of Factorials
Factorials grow very rapidly. They appear frequently in combinatorics, probability, and calculus.
- For n \(\ge\) k, n! is always divisible by k!.
- For n \(\ge\) prime number p, n! is always divisible by p. This is because p will be one of the factors in the product \( n \times (n-1) \times \dots \times 1 \).
- For n \(\ge\) composite number c, n! is always divisible by c, provided c can be expressed as a product of prime factors, and each prime factor (with its power) is less than or equal to its highest power in the prime factorization of n!. For example, for n \(\ge\) 6, n! is divisible by 6 (since \( 6 = 2 \times 3 \), and both 2 and 3 are factors within \( 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 \)). For n \(\ge\) 4, n! is divisible by 4 (since \( 4 = 2 \times 2 \), and \( 4! = 24 \) contains \( 2^3 \)).
- Wilson's Theorem states that for a prime number p, \( (p-1)! \equiv -1 \pmod{p} \). This relates factorials to prime numbers using modular arithmetic.
Understanding the properties of factorials and divisibility rules is crucial for solving problems like the one discussed here.
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