Introduction to Finding Remainders Using Modular Arithmetic

When dealing with large exponents like 2^{2015}, it can be challenging to compute the remainder when divided by a smaller number, such as 3. In this article, we will explore various methods to determine this remainder, including the application of modular arithmetic and Fermat's Little Theorem. By the end of this guide, you will have a solid understanding of how to solve such problems efficiently.

1. Understanding the Problem

The objective is to find the remainder when 2^{2015} is divided by 3. This problem can be approached using several techniques, each offering a different perspective on the solution.

2. Application of Modular Arithmetic

Let's start by examining the pattern of 2^n mod 3:

2^1 ≡ 2 mod 3 2^2 ≡ 4 ≡ 1 mod 3 2^3 ≡ 2 * 2^2 ≡ 2 * 1 ≡ 2 mod 3 2^4 ≡ 2 * 2^3 ≡ 2 * 2 ≡ 4 ≡ 1 mod 3

From this pattern, we observe that the powers of 2 modulo 3 alternate between 2 and 1. Therefore, if the exponent is odd, the remainder is 2, and if the exponent is even, the remainder is 1. Since 2015 is an odd number, the remainder of 2^{2015} mod 3 is 2.

Answer: The remainder is boxed{2}.

3. Using Negative Equivalents in Modular Arithmetic

An alternative approach is to use the fact that 2 ≡ -1 mod 3. This means that we can replace 2 with -1 in the expression:

2^{2015} ≡ (-1)^{2015} mod 3 ≡ -1 mod 3 ≡ 2 mod 3

Because -1 ≡ 2 mod 3, the remainder is again 2.

Answer: The remainder is boxed{2}.

4. Application of 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} ≡ 1 mod p

In our case, p 3, and a 2. Thus:

2^{3-1} ≡ 2^2 ≡ 1 mod 3

Using this, we can write:

2^{2015} 2^{2 * 1007 1} (2^2)^{1007} * 2 ≡ 1^{1007} * 2 ≡ 2 mod 3

So, the remainder is boxed{2}.

5. Generalization of the Pattern

A more general insight is that the remainder when 2 is raised to an odd power and divided by 3 is always 2, and for even powers, it is always 1. This can be seen through the patterns observed in the powers of 2 modulo 3:

2^1 ≡ 2 mod 3

2^2 ≡ 1 mod 3

2^3 ≡ 2 mod 3

2^4 ≡ 1 mod 3

and so on.

By understanding this pattern, we can efficiently compute the remainder for any power of 2 without having to perform the full exponentiation.

Conclusion

Using modular arithmetic, the negative equivalent method, and Fermat's Little Theorem, we have demonstrated multiple ways to find the remainder when 2^{2015} is divided by 3. The remainder is consistently found to be 2. These techniques are not only useful for this specific problem but also form a foundational understanding of modular arithmetic, which is applicable across a wide range of mathematical and computational contexts.