Problem 98 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If 7 divides \(32^{3211}\), the remainder is (A) 2 (B) 4 (C) 8 (D) none of these

Step-by-Step Solution

Answer

(A) 2

Step 1: Understanding the Problem

The problem asks us to find the remainder when \(32^{3211}\) is divided by 7. This can be solved by using the properties of modular arithmetic.

Step 2: Simplifying the Base Modulo 7

First, we simplify the base 32 modulo 7. Divide 32 by 7, we get 32 \(\equiv\) 4 (mod 7), because 32 divided by 7 leaves a remainder of 4.

Step 3: Applying Fermat's Little Theorem

Fermat's Little Theorem states that if \(p\) is a prime number and \(a\) is an integer not divisible by \(p\), then \(a^{p-1} \equiv 1\) (mod \(p\)). Here, \(a = 4\) and \(p = 7\), so \(4^6 \equiv 1\) (mod 7).

Step 4: Reducing the Exponent

Now, reduce the exponent 3211 in \(4^{3211}\) using the fact 3211 mod 6 equals 5 (since 3211 divided by 6 gives a remainder of 5). Hence, \(4^{3211} \equiv 4^5\) (mod 7).

Step 5: Calculating \(4^5\) Modulo 7

Calculate \(4^5\). We find that \(4^5 = 1024\). Now, calculate \(1024\) mod 7. Divide 1024 by 7, the remainder is 2. Thus, \(4^5 \equiv 2\) (mod 7).

Step 6: Concluding the Answer

The remainder when \(32^{3211}\) is divided by 7 is 2. Therefore, the correct answer choice is (A) 2.

Key Concepts

Fermat's Little Theorem

Fermat's Little Theorem is a foundational concept in number theory and modular arithmetic. It helps simplify complex calculations involving powers and remainders, which appear in exercises like the one provided. The theorem states that if \( p \) is a prime number and \( a \) is an integer not divisible by \( p \), then:

• \( a^{p-1} \equiv 1 \pmod{p} \)

This statement means if you raise the integer \( a \) to the power of one less than \( p \) and then divide by \( p \), the remainder is always 1. This surprising property works because it simplifies exponentiation in modulo arithmetic.
In the example provided, 4 is raised to a power, and since studying powers of numbers is often complex, we apply Fermat's theorem to simplify the computation process. By knowing that \( 4^6 \equiv 1 \pmod{7} \), we reduce the problem's complexity by recognizing repetitive cycles in the powers of 4.

Exponentiation Modulo

Exponentiation modulo addresses the calculation of large powers with respect to their remainders when divided by a number. In simpler terms, it is the operation of finding what a large power of a number equals when divided by another number until there is no remainder. This is particularly useful in computer science and cryptography.
For example, consider calculating \( 32^{3211} \mod 7 \). Direct exponentiation would be computationally impractical due to the very large size of the power. Instead, we use smaller steps, reducing bases and exponents using modular properties:

• Simplify the base first, such as finding \( 32 \equiv 4 \pmod{7} \).
• Apply rules like Fermat's Little Theorem to reduce the exponent size, as shown by reducing \( 3211 \mod 6 \) to simplify calculations to just \( 4^5 \).
• Finally, using smaller mod calculations like finding \( 1024 \mod 7 \) to conclude with the remainder.

This method makes handling large numbers simpler and remains efficient, avoiding extensive and potentially error-prone direct calculations.

Remainder Theorem

The Remainder Theorem plays a crucial role in modular arithmetic, especially when dividing numbers and understanding the notion of 'remainders' better. When a number, called the dividend, is divided by another number, the divisor, we talk about how much is left over after the division process, which is called the remainder.
When solving problems involving modular arithmetic, as seen in the given exercise, this theorem lets us find straightforward solutions for what would otherwise be complicated polynomial calculations. It involves:

• Understanding how numbers behave under division and finding similar patterns.
• Using patterns and properties, like those in Fermat's Little Theorem, to simplify solutions.
• Applying these patterns so that instead of complex calculations, you break down tasks into finding simple remainders.

The exercise here shows how recognizing a remainder beforehand—like simplifying \( 32 \equiv 4 \pmod{7} \)—can make solving seemingly large exponentiation tasks much more accessible and manageable by using essential modular arithmetic principles.