Problem 130 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(n

Step-by-Step Solution

Answer

(D) \(p^{2}\) does not divide \(N\).

Step 1: Understanding the Binomial Coefficient

The symbol \( {}^{2n}C_n \) represents the binomial coefficient given by \( \frac{(2n)!}{n!n!} \). This is the number of ways to choose \( n \) items from \( 2n \) items, and is also known as the central binomial coefficient.

Step 2: Identify the Problem Constraints

We have that \( n < p < 2n \) and \( p \) is a prime number that may or may not divide \( N \). We need to determine how \( p \) relates to \( N \), which is equal to \( {}^{2n}C_n \).

Step 3: Apply Lucas' Theorem

Lucas’ Theorem provides a tool to determine divisibility of binomial coefficients by a prime. It implies that \( {}^{2n}C_n \equiv 0 \pmod{p} \) unless the binomial coefficients of the digits of \( 2n \) and \( n \) when expressed in base \( p \) are less than \( p \). Given \( n < p < 2n \), at least one digit of \( n \) in base \( p \) must be non-zero, implying \( {}^{2n}C_n \equiv 0 \pmod{p}\). Thus, \( p \mid N \).

Step 4: Consider Divisibility by \( p^2 \)

Even though \( p \mid N \), for \( p^2 \mid N \), we need a deeper exploration into factorials in divide \( N = \frac{(2n)!}{(n!)^2} \). Typically, \( p \) being a prime between \( n \) and \( 2n \) means \( p \) occurs once within \( [(2n)!] \) divided by \( [(n!)]^2 \). Hence, \( p^2 \) does not divide \( N \).

Step 5: Final Conclusion

Since \( p \mid N \) but \( p^2 mid N \), the correct choice from the options given is that \( p^2 \) does not divide \( N \).

Key Concepts

Prime Numbers

Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves. Understanding prime numbers is essential in many areas of mathematics, including number theory and combinatorics.

Here are some key points about prime numbers:

• Every number is either a prime or a product of primes.
• Primes are used as building blocks for other numbers, a concept known as prime factorization.
• They have unique properties that make them crucial in cryptography and secure communications.

Prime numbers only ensure divisibility in specific scenarios within mathematical problems, influencing results like the binomial coefficient's divisibility.

In the given exercise, the prime number, denoted as \(p\), plays a central role in determining the divisibility of the binomial coefficient \(^{2n}C_n\), hinting at profound connections between basic arithmetic properties and complex mathematical theorems.

Lucas' Theorem

Lucas' Theorem is a fascinating insight into divisibility concerning binomial coefficients.It specifies conditions under which certain binomial coefficients are divisible by a number.

The theorem provides:

• A method to reduce the problem of determining divisibility by a prime to examining smaller coefficients.
• Utilization of base \(p\) representation to understand the relationship between numbers.
• A powerful tool for tackling problems involving large numbers or high powers.

In the exercise, Lucas' Theorem helps us ascertain whether \(p\) divides the binomial coefficient without direct computation.
Given \(n < p < 2n\) and knowing \( {}^{2n}C_n \equiv 0 \pmod{p} \), Lucas' Theorem indicates that \(p\) divides \(^{2n}C_n\).

This simplifies verifying divisibility properties conveniently and supports the analytical exploration involved in the solution.

Divisibility

Divisibility, a foundational concept in mathematics, pertains to how numbers divide each other without leaving remainders.It's essential in simplifying expressions and solving equations, and it plays a crucial part in understanding number properties.

Key aspects include:

• Determining if one number is a factor of another.
• Applying divisibility rules to quickly check for factors like 2, 3, 5, etc.
• Essential in understanding congruences and modular arithmetic, as used in this problem.

In the exercise, divisibility drives the central question regarding whether a prime number \( p \) or its power \( p^2 \) divides the binomial coefficient.

The analysis shows that while \( p \) divides \(N\), \( p^2 \) does not.This insight, informed through examining how primes integrate into factorial structures, helps us conclude that \( p^2 \) indeed does not divide \(N\).