Problem 99 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The largest integer \(n\) such that \(33 !\) is divisible by \(2^{n}\) is (A) 30 (B) 31 (C) 32 (D) None of these

Step-by-Step Solution

Answer

The answer is (B) 31.

Step 1: Understanding Factorial and Divisibility

We need to find the largest power of 2 that divides the factorial of 33. The factorial of 33, denoted as \(33!\), is the product of all positive integers up to 33. To determine the highest power of 2 that divides this product, we use the formula for the largest power of a prime \(p\) dividing \(n!\): \(\sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor\). In this case, \(p = 2\) and \(n = 33\).

Step 2: Applying the Formula

Let's apply the formula to find the largest power of 2 dividing \(33!\): \[ \left\lfloor \frac{33}{2} \right\rfloor + \left\lfloor \frac{33}{4} \right\rfloor + \left\lfloor \frac{33}{8} \right\rfloor + \left\lfloor \frac{33}{16} \right\rfloor + \left\lfloor \frac{33}{32} \right\rfloor. \]

Step 3: Calculating Each Component

Calculate each component of the sum:1. \(\left\lfloor \frac{33}{2} \right\rfloor = \left\lfloor 16.5 \right\rfloor = 16\)2. \(\left\lfloor \frac{33}{4} \right\rfloor = \left\lfloor 8.25 \right\rfloor = 8\)3. \(\left\lfloor \frac{33}{8} \right\rfloor = \left\lfloor 4.125 \right\rfloor = 4\)4. \(\left\lfloor \frac{33}{16} \right\rfloor = \left\lfloor 2.0625 \right\rfloor = 2\)5. \(\left\lfloor \frac{33}{32} \right\rfloor = \left\lfloor 1.03125 \right\rfloor = 1\)

Step 4: Summing Components

Add up all the calculated values: \(16 + 8 + 4 + 2 + 1 = 31\). Thus, the largest integer \(n\) such that \(33!\) is divisible by \(2^n\) is 31.

Key Concepts

Factorial

A factorial, represented as \( n! \), is the product of all positive integers from 1 to \( n \). For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \). Factorials grow very quickly as \( n \) increases.
Understanding factorials is crucial in various mathematical problems, especially those involving permutations and combinations. For our problem, we need to calculate \( 33! \), but we don't need the actual number. Instead, we're interested in how many times a certain number, like 2, is a factor in the complete product of numbers up to 33. This understanding lays the groundwork for exploring problems involving prime factorization and divisibility.

Prime Factorization

Prime factorization involves expressing a number as a product of prime numbers. A prime number is a whole number greater than 1, only divisible by 1 and itself.
For example, the number 18 can be factored into primes as: \( 18 = 2 \times 3^2 \). This method is useful in various applications like finding least common multiples or greatest common divisors. In our context, we're dissecting the factorial of 33 to determine how many times the number 2 appears in its prime factorization. Using the formula for the highest power of a prime \( p \) dividing \( n! \), we systematically calculate the exact power of 2 present in the expansion of \( 33! \).

• Calculate each \( \left\lfloor \frac{n}{p^k} \right\rfloor \) where \( p = 2 \).
• Add these values to find how many 2's divide perfectly into \( 33! \).

This approach makes it clear when evaluating the power of any prime number in a factorial.

Mathematical Theorems

Mathematical theorems provide proven statements that help us solve problems or derive other truths easily. In this context, we're utilizing a theorem related to the prime factorization of factorials: if you want to know the largest power of a prime \( p \) dividing \( n! \), use \( \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \).
These mathematical steps become straightforward via this formula, allowing rapid calculations that would otherwise be time-consuming. The steps involve systematically reducing the fraction, rounding down, and summing these values to get our result.

• This process is invaluable for tasks in combinatorics and number theory.
• It shows the power of rigor and logic in mathematics, enabling us to handle large numbers and complex computations reliably.

Floor Function

The floor function, denoted as \( \left\lfloor x \right\rfloor \), is an important mathematical operation. It outputs the greatest integer less than or equal to \( x \). For instance, \( \left\lfloor 4.9 \right\rfloor = 4 \) and \( \left\lfloor 7 \right\rfloor = 7 \).
This function is particularly helpful in divisibility problems, as it simplifies the calculations involving non-integers. In our solution, the floor function helps us truncate decimals when computing powers of a factor in a factorial.

• Ensures whole number results when dealing with partial factors.
• Allows precise calculations without estimations or rounding errors.

This piece of mathematics aids in breaking down problems step by step, especially when integrating factors into larger calculations.