Problem 3 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\\{x\\}\) denotes the fractional part of \(x\), then \(\left\\{\frac{2^{2003}}{17}\right\\}\) is \(\begin{array}{ll}\text { (A) } 2 / 17 & \text { (B) } 4 / 17\end{array}\) (C) \(8 / 17\) (D) \(16 / 17\)

Step-by-Step Solution

Answer

The answer is (C) \(8/17\).

Step 1: Understand the Fractional Part Function

The fractional part of a number \(x\), denoted as \(\{x\}\), is defined as \(x - \lfloor x \rfloor\). It represents the part of \(x\) that comes after the decimal point.

Step 2: Compute the Expression

Given the expression \(\left\{ \frac{2^{2003}}{17} \right\}\), we need to evaluate \( 2^{2003} \mod 17 \). This is because \(\left\{ \frac{2^{2003}}{17} \right\}\) is equivalent to \(\frac{2^{2003} \mod 17}{17}\).

Step 3: Apply Fermat's Little Theorem

Fermat's Little Theorem tells us that if \(p\) is a prime number, then for any integer \(a\) not divisible by \(p\), \(a^{p-1} \equiv 1 \pmod{p}\). Here, \(a = 2\) and \(p = 17\), so \(2^{16} \equiv 1 \pmod{17}\).

Step 4: Simplify the Exponent

Since we want \(2^{2003}\), we reduce the exponent modulo 16 (since \(2^{16} \equiv 1 \pmod{17}\)). Calculate \(2003 \mod 16\) to find this reduced exponent.

Step 5: Calculate the Reduced Exponent

Perform the division: \(2003 \div 16 = 125\) with a remainder of \(3\). Hence, \(2003 \equiv 3 \pmod{16}\).

Step 6: Compute 2 to the Reduced Exponent

Therefore, \(2^{2003} \equiv 2^3 \pmod{17}\). Calculate \(2^3 = 8\), and thus \(2^{2003} \equiv 8 \pmod{17}\).

Step 7: Determine the Fractional Part

Since \(2^{2003} \equiv 8 \pmod{17}\), \(\left\{ \frac{2^{2003}}{17} \right\} = \frac{8}{17}\). Thus, the fractional part is the fraction \(\frac{8}{17}\).

Key Concepts

Fermat's Little Theorem

Fermat's Little Theorem is a fundamental principle in number theory that provides an easy way to work with powers of numbers in modular arithmetic. It is named after the mathematician Pierre de Fermat. This theorem states that for any integer \(a\) and a prime number \(p\), the congruence \(a^{p-1} \equiv 1 \pmod{p}\) holds true, provided that \(a\) is not divisible by \(p\).

This powerful theorem simplifies calculations involving large exponents by reducing them to a manageable size. For example, to solve within a modulo context like \(2^{2003} \mod 17\), Fermat's Little Theorem tells us that since \(17\) is prime, \(2^{16} \equiv 1 \pmod{17}\). Knowing this, we can greatly simplify calculations of larger exponents such as \(2003\) using a modulo of \(16\). By breaking down these numbers into smaller congruent parts, the theorem helps manage calculations that would otherwise be extremely cumbersome.

Modulo Operations

In mathematics, modulo operations are a method of finding the remainder when one number is divided by another. It is often represented with the percentage symbol, \(\%\), or simply as \(\mod\). The result of a modulo operation is always less than the divisor. Thus, if we say \(a \bmod b = r\), \(r\) is the remainder when \(a\) is divided by \(b\), and \(0 \leq r < b\).

In the context of the given problem, we need to calculate \(2^{2003} \mod 17\). By doing this, we effectively reduce the large number \(2^{2003}\) to something much simpler, without losing the essence of our calculations. This technique is especially useful for computations in cryptography and in computer science, where handling smaller numbers leads to efficiency. In this exercise, we found that \(2^{2003} \equiv 8 \pmod{17}\), which is a much simpler expression to evaluate.

Exponents in Number Theory

Exponents in number theory are pivotal in understanding growth rates, patterns, and solving various equations. In number theory, managing large exponents is crucial, especially within the context of modulo operations. By applying principles such as Fermat's Little Theorem, we can simplify expressions involving large powers.

For instance, handling \(2^{2003}\) directly would be infeasible due to its size. Instead, we use modular arithmetic to reduce the exponent modulo a smaller number. As demonstrated in the problem, knowing that \(2^{16} \equiv 1 \pmod{17}\), we reduce \(2003\) to a smaller exponent using \(2003 \mod 16\), which yields \(3\). This reduction is essential as it transforms a huge problem into a manageable one: \(2^3 = 8\), giving us \(2^{2003} \equiv 8 \pmod{17}\). Using these techniques greatly simplifies number theory problems and broadens our ability to solve complex equations efficiently.