2^50 Divided By 7: Finding The Remainder Explained

Hey there, math enthusiasts! Ever found yourself scratching your head over seemingly complex division problems? Well, today we're going to unravel a fascinating one: What's the remainder when 2^50 is divided by 7? It might sound intimidating, but trust me, we'll break it down into manageable steps. We will explore the beauty of modular arithmetic and some cool theorems that make this challenge a piece of cake. Get ready to flex those mathematical muscles! So, let’s dive into the exciting world of numbers and remainders.

The Challenge: 2^50 Modulo 7

Okay, let's lay out the problem clearly. We're trying to find the remainder when 2^50 is divided by 7. In mathematical terms, we want to compute 2^50 mod 7. Now, you might be thinking, “Do I really need to calculate 2^50 and then divide it by 7?” Thankfully, no! That would be a monstrous task. Instead, we can use some clever techniques to simplify this. We'll explore Fermat's Little Theorem and the concept of modular exponentiation. These tools are like secret weapons in the world of number theory. They allow us to tackle big problems without getting bogged down in huge calculations. So, buckle up, because we're about to embark on a journey to discover the remainder of this seemingly complex problem.

Option Analysis: A Sneak Peek

Before we jump into the calculations, let's take a quick look at the multiple-choice options. We have:

• A) 1
• B) 2
• C) 4
• D) 5

Knowing the possible answers can sometimes give us a hint or help us check our work later. It’s like having a roadmap before a journey – you know where you’re headed. Now, let's keep these options in mind as we delve into the methods to solve this problem. This will help us stay focused and ensure we arrive at the correct answer. Remember, in math, every little detail counts!

Method 1: Fermat's Little Theorem – A Quick Solution

Let's start with one of the most elegant tools in number theory: Fermat's Little Theorem. This theorem is a gem when dealing with remainders and prime numbers. It states that if p is a prime number, then for any integer a not divisible by p, a^(p-1) ≡ 1 (mod p). In simpler words, if you raise a number a to the power of (p-1) and divide it by p, the remainder will be 1. Now, how does this apply to our problem? Well, 7 is a prime number, and 2 is not divisible by 7. So, according to Fermat's Little Theorem:

2^(7-1) ≡ 2^6 ≡ 1 (mod 7)

This is a crucial piece of information. It tells us that 2^6 leaves a remainder of 1 when divided by 7. Now, we can use this to simplify 2^50. Think of it like this: we've found a repeating pattern in the remainders of powers of 2 modulo 7. By understanding this pattern, we can break down the large exponent of 50 into smaller, more manageable chunks. This is the essence of modular arithmetic – finding patterns and using them to our advantage. So, let's see how we can apply this newfound knowledge to solve our original problem.

Applying Fermat's Little Theorem

Now that we know 2^6 ≡ 1 (mod 7), we can rewrite 2^50 in terms of 2^6. We can express 50 as a multiple of 6 plus a remainder:

50 = 6 * 8 + 2

So, we can rewrite 2^50 as:

2^50 = 2^(6*8 + 2) = (2⁶⁾8 * 2^2

Now, using our knowledge of modular arithmetic, we can say:

(2⁶⁾8 * 2^2 ≡ (1)^8 * 2^2 (mod 7)

Since 1 raised to any power is still 1, this simplifies to:

1 * 2^2 ≡ 4 (mod 7)

And there you have it! The remainder when 2^50 is divided by 7 is 4. This is the power of Fermat's Little Theorem – it allows us to drastically simplify complex calculations by finding patterns in modular arithmetic. It's like finding a shortcut through a maze. We started with a seemingly daunting problem, but by applying this theorem, we arrived at the solution with relative ease. So, the answer is C) 4. Let’s move on to the next method to confirm our result.

Method 2: Modular Exponentiation – Step-by-Step

Let's explore another method to solve this problem: modular exponentiation. This technique involves breaking down the exponent and calculating remainders step by step. It's a bit more hands-on than Fermat's Little Theorem, but it provides a solid understanding of how remainders work with exponents. We'll calculate the remainders of smaller powers of 2 when divided by 7, and then use these results to find the remainder of 2^50. Think of it as building a staircase – each step gets us closer to the final answer. This method is particularly useful when you need to calculate the remainder of a large power modulo a number, and it's a fundamental concept in computer science and cryptography.

Step-by-Step Calculation

We'll start by calculating the first few powers of 2 modulo 7:

• 2^1 ≡ 2 (mod 7)
• 2^2 ≡ 4 (mod 7)
• 2^3 ≡ 8 ≡ 1 (mod 7)

Ah, we've found a key pattern here! 2^3 leaves a remainder of 1 when divided by 7. This is similar to what we found with Fermat's Little Theorem, but this time we discovered it through direct calculation. This pattern is crucial because it allows us to simplify the calculation of 2^50. Just like before, we can rewrite 2^50 in a way that utilizes this pattern. It's like finding a secret code that unlocks the solution. Now, let's use this pattern to break down 2^50 and find its remainder when divided by 7.

Applying the Pattern

Since 2^3 ≡ 1 (mod 7), we can rewrite 2^50 using powers of 3 in the exponent. We can express 50 as a multiple of 3 plus a remainder:

50 = 3 * 16 + 2

So, we can rewrite 2^50 as:

2^50 = 2^(3*16 + 2) = (2³⁾16 * 2^2

Now, using our modular arithmetic knowledge:

(2³⁾16 * 2^2 ≡ (1)^16 * 2^2 (mod 7)

Since 1 raised to any power is still 1, this simplifies to:

1 * 2^2 ≡ 4 (mod 7)

Again, we arrive at the same answer: the remainder when 2^50 is divided by 7 is 4. This confirms our result from Method 1, strengthening our confidence in the solution. Modular exponentiation is a powerful technique that allows us to tackle large exponents by breaking them down into smaller, more manageable steps. It's like climbing a mountain one step at a time – eventually, you reach the summit.

Conclusion: The Remainder Revealed

So, after exploring two different methods – Fermat's Little Theorem and modular exponentiation – we've confidently arrived at the answer. The remainder when 2^50 is divided by 7 is indeed 4. We've seen how these techniques can simplify seemingly complex problems and reveal the hidden beauty of number theory. It's like being a mathematical detective, using clues and tools to solve a mystery. Remember, math is not just about numbers and formulas; it's about problem-solving, logical thinking, and the joy of discovery.

Final Answer and Justification

The correct answer is C) 4. We've demonstrated this using both Fermat's Little Theorem and modular exponentiation. Fermat's Little Theorem provided a quick and elegant solution by utilizing the property that 2^6 ≡ 1 (mod 7). Modular exponentiation, on the other hand, offered a step-by-step approach, reinforcing our understanding of how remainders work with exponents. Both methods highlight the power of modular arithmetic in simplifying complex calculations. It's like having two different keys that unlock the same door. By using these techniques, we can tackle similar problems with confidence and precision. So, the next time you encounter a seemingly daunting division problem, remember the tools and techniques we've discussed today, and you'll be well on your way to finding the solution.