Problem 1 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \(\\{p\\}\) denotes the fractional part of the number \(p\), then \(\left\\{\frac{3^{200}}{8}\right\\}\), is equal to: (a) \(\frac{5}{8}\) (b) \(\frac{7}{8}\) (c) \(\frac{3}{8}\) (d) \(\frac{1}{8}\)

Step-by-Step Solution

Answer

The fractional part is \(\frac{1}{8}\), so the answer is (d).

Step 1: Understanding the Fractional Part

The fractional part of a number \(p\), denoted as \(\{p\}\), is given by \(p - \lfloor p \rfloor\), where \(\lfloor p \rfloor\) is the greatest integer less than or equal to \(p\). This exercise asks us to find the fractional part of \(\frac{3^{200}}{8}\).

Step 2: Modular Arithmetic Approach

To find \(\left\{ \frac{3^{200}}{8} \right\}\), we start by considering \(3^{200}\) modulo 8. We will use Fermat's Little Theorem to simplify our calculations since 8 is not a prime, but we can consider powers of 3 modulo 8.

Step 3: Calculating Power Remainders

By calculating, we find: \(3^1 \equiv 3 \pmod{8}\), \(3^2 \equiv 1 \pmod{8}\). Therefore, for any even power \(k\), \(3^{2k} \equiv 1 \pmod{8}\). Specifically, \(3^{200} = (3^2)^{100} \equiv 1^{100} \equiv 1 \pmod{8}\).

Step 4: Determining Fractional Part

Since \(3^{200} \equiv 1 \pmod{8}\), we can write \(3^{200} = 8k + 1\) for some integer \(k\). Thus, \(\frac{3^{200}}{8} = k + \frac{1}{8}\). The fractional part, hence, is \(\frac{1}{8}\).

Step 5: Answer Verification

Given the choices (a) \(\frac{5}{8}\), (b) \(\frac{7}{8}\), (c) \(\frac{3}{8}\), (d) \(\frac{1}{8}\), the correct answer is \(\frac{1}{8}\), which matches our calculation of the fractional part.

Key Concepts

Fractional Part

The fractional part of any number helps us understand the decimal remaining after the whole number part is taken away. In mathematics, the fractional part of a number \( p \), denoted by \( \{p\} \), is found by subtracting the largest whole number less than or equal to \( p \) from \( p \) itself. For example:

• If you have \( 3.75 \), the fractional part is \( 0.75 \), because \( 3.75 - 3 = 0.75 \).
• For a negative number like \( -2.5 \), the fractional part is \( 0.5 \), because it is effectively calculated as \( -2.5 - (-3) = 0.5 \).

Understanding fractional parts is useful in many areas of mathematics, especially when dealing with cyclic or periodic phenomena, as it helps isolate the 'leftover' component after full cycles are accounted for. This concept is crucial in many mathematical exercises, including the one where you need to compute \( \left\{ \frac{3^{200}}{8} \right\} \), as it involves identifying the remainder portion of a division.

Modular Arithmetic

Modular arithmetic is like clock arithmetic; you're working with remainders. It is key in simplifying calculations involving large powers, such as in the case of finding the fractional part of \( \frac{3^{200}}{8} \). Essentially, when we say \( a \equiv b \pmod{n} \), we mean that \( a \) and \( b \) give the same remainder when divided by \( n \).

This concept allows us to break down large exponentiations into simpler forms. For instance, without calculating a gigantic number like \( 3^{200} \), you can find its remainder when divided by 8 by:

• First calculating smaller powers of 3 up to the modulus: \( 3^1 \equiv 3 \pmod{8} \) and \( 3^2 \equiv 1 \pmod{8} \).
• Then using these results to conclude that subsequent even powers such as \( 3^{200} = (3^2)^{100} \equiv 1^{100} \equiv 1 \pmod{8} \).

Thus, modular arithmetic can simplify complex problems by reducing them to easily manageable calculations.

Fermat's Little Theorem

Fermat's Little Theorem is a fundamental principle in number theory that helps efficiently calculate large powers modulo a prime number. The theorem states: if \( p \) is a prime and \( a \) is an integer not divisible by \( p \), then \( a^{p-1} \equiv 1 \pmod{p} \). Even though 8 is not a prime number, understanding this theorem can improve your comprehension of modular arithmetic techniques when solving similar types of problems.

• In the context of our problem, it guides you to look for patterns like \( a^k \equiv 1 \pmod{8} \) where \( k \) could be a smaller sequence, simplifying calculations.
• For example, the realization that \( 3^2 \equiv 1 \pmod{8} \) effectively shows how powerful this simplification tool can be: it allows calculating powers like \( 3^{200} \equiv (3^2)^{100} \equiv 1^{100} \equiv 1 \pmod{8} \), following a similar pattern recognition approach inspired by Fermat's principle.

While the theorem doesn’t apply directly to non-prime numbers, its underlying strategy helps solve many modular arithmetic problems involving large powers.