Problem 20 - 71 JEE Main Mathematics Online (2020 - 2012) & Offline (2018 - 2002) Chapterwise + Topicwise Solved Papers

Question

If \((27)^{999}\) is divided by 7 , then the remainder is : (a) 1 (b) 2 (c) 3 (d) 6

Step-by-Step Solution

Answer

The remainder is 6.

Step 1: Understanding the Problem

The task is to find the remainder of \((27)^{999}\) when divided by 7. We need to determine the value of this expression modulo 7.

Step 2: Use Modular Arithmetic

First, simplify \(27\mod 7\). We calculate: \(27 \div 7 = 3\) remainder 6. Thus, \(27 \equiv 6 \mod 7\). Then, the problem simplifies to finding \((6)^{999} \mod 7\).

Step 3: Apply Fermat's Little Theorem

Fermat's Little Theorem states that \(a^{p-1} \equiv 1 \mod p\) for any integer \(a\) that is not divisible by a prime \(p\). Here, \(p = 7\), so \(6^6 \equiv 1 \mod 7\). We first check this: \(6^6 = 46656\). Calculating \(46656\mod 7\) gives a remainder of 1, so this confirms that \(6^6 \equiv 1 \mod 7\).

Step 4: Divide the Exponent by 6

Now, decompose the exponent 999 into multiples of 6. We calculate: \(999 \div 6 = 166\) remainder 3. Thus, \(999 = 6\times 166 + 3\), allowing us to use the result from step 3.

Step 5: Simplify the Expression

Using the decomposition, we have: \(6^{999} = (6^6)^{166} \times 6^3\). Because \(6^6 \equiv 1 \mod 7\), \((6^6)^{166} \equiv 1 \mod 7\). Therefore, \(6^{999} \equiv 6^3 \mod 7\).

Step 6: Calculate the Final Remainder

Now, compute \(6^3 = 216\). Finding \(216 \mod 7\), \(216 \div 7 = 30\) remainder 6. Hence, \(6^3 \equiv 6 \mod 7\). Therefore, the remainder when \((27)^{999}\) is divided by 7 is 6.

Key Concepts

Fermat's Little Theorem

Fermat's Little Theorem is a fundamental result in number theory that helps in simplifying calculations involving exponents in modular arithmetic. This theorem states that if you have a prime number \(p\) and any integer \(a\) that is not divisible by \(p\), then it follows that \(a^{p-1} \equiv 1 \mod p\).
For example, in the given problem, we applied Fermat's Little Theorem to simplify \(6^{999} \mod 7\). Here 7 is prime, and since 6 is not divisible by 7, we use the theorem to conclude that \(6^6 \equiv 1 \mod 7\).
Using Fermat's Little Theorem in this way allows us to reduce large exponentiation calculations into smaller, more manageable ones. This step is particularly helpful when dealing with very large powers like in our original problem.

Exponentiation

Exponentiation is the mathematical process of raising one number (the base) to the power of another number (the exponent). In modular arithmetic, exponentiation becomes more complex, especially with large exponents.
For example, calculating \(27^{999}\) directly would be computationally intensive. However, by using the properties of modular arithmetic, which includes reducing the base modulo the divisor before exponentiating, we simplify the problem.
We reduced \(27\) to \(6\) modulo 7, transforming the problem from \(27^{999} \mod 7\) to \(6^{999} \mod 7\). Efficient strategies like this help manage large exponentiation problems in modular arithmetic applications.

Remainder Theorem

The Remainder Theorem in modular arithmetic is straightforward - it determines the remainder of a number when it is divided by another number. It is a crucial concept for simplifying calculations and is central to modular arithmetic.
In the exercise, we used this concept at multiple stages. First, to reduce 27 modulo 7, resulting in a simpler problem of \(6^{999} \mod 7\). Later, after applying Fermat's Little Theorem and simplifying the expression, we found the remainder of \(6^3\) when divided by 7.
This step-by-step utilization of the Remainder Theorem helps in verifying calculations and ensuring accuracy in modular arithmetic problems. It's imperative for students to practice this technique to solve complex problems effectively.