Finding The Remainder Of 4^{103} + 4^{104} + 4^{105} Divided By 13

Understanding remainders in mathematics is a crucial skill, particularly when dealing with large exponents and modular arithmetic. This article explores a specific problem: determining the remainder when the sum of three exponential terms, 4^{103} + 4^{104} + 4^{105}, is divided by 13. We'll break down the problem step-by-step, utilizing properties of exponents and modular arithmetic to arrive at the solution. Our journey will involve simplifying the expression, identifying patterns in powers of 4 modulo 13, and ultimately calculating the final remainder. This exploration not only provides the answer to this specific question but also enhances our understanding of more general techniques applicable to similar problems in number theory.

Simplifying the Expression

To begin, let's simplify the given expression: 4^{103} + 4^{104} + 4^{105}. Notice that each term has a common factor of 4^{103}. We can factor this out to make the expression easier to handle. Factoring out common terms is a fundamental algebraic technique that simplifies expressions and reveals underlying structures. In this specific case, factoring out 4^{103} transforms the expression from a sum of individual exponential terms into a product of an exponential term and a simpler sum. This approach is a cornerstone of problem-solving in mathematics, particularly when dealing with complex expressions. By strategically identifying and factoring out common factors, we can often reduce a seemingly daunting problem into more manageable parts. This initial simplification is a crucial step towards solving the remainder problem. When dealing with large exponents, simplifying the base expression can significantly reduce computational complexity and reveal patterns that are otherwise hidden. Let's see how this simplification unfolds and paves the way for finding the remainder.

Factoring out 4^{103}, we get:

4^{103} + 4^{104} + 4^{105} = 4^{103}(1 + 4^1 + 4^2) = 4^{103}(1 + 4 + 16) = 4^{103}(21).

Now we need to find the remainder when 4^103}(21) is divided by 13. The simplification process reduces the problem to finding the remainder of a product, which can be further analyzed using modular arithmetic properties. Remember that the key idea here is that if we can find the remainders of the individual factors when divided by 13, we can multiply those remainders to find the remainder of the entire product. The simplification we performed earlier allows us to separate the original expression into two manageable factors 4^{103 and 21. This approach is particularly useful when dealing with large exponents, as directly computing 4^{103} would be computationally intensive. By focusing on the remainders, we can work with smaller numbers and exploit the cyclical patterns that often emerge in modular arithmetic. This step is crucial because it transforms a complex calculation into a series of simpler ones, each of which is easier to handle. Let's continue to explore how modular arithmetic will help us crack this problem.

Modular Arithmetic and Powers of 4

Modular arithmetic deals with remainders after division. We write a ≡ b (mod m) if a and b have the same remainder when divided by m. To find the remainder when 4^{103}(21) is divided by 13, we can first find the remainders of 4^{103} and 21 separately when divided by 13, and then multiply those remainders. This is a fundamental property of modular arithmetic: (a * b) mod m = ((a mod m) * (b mod m)) mod m.

First, let's find the remainder when 21 is divided by 13. 21 divided by 13 gives a quotient of 1 and a remainder of 8. So, 21 ≡ 8 (mod 13).

Next, we need to find the remainder when 4^{103} is divided by 13. To do this, let's look at the powers of 4 modulo 13:

• 4^1 ≡ 4 (mod 13)
• 4^2 ≡ 16 ≡ 3 (mod 13)
• 4^3 ≡ 4 * 3 ≡ 12 (mod 13)
• 4^4 ≡ 4 * 12 ≡ 48 ≡ 9 (mod 13)
• 4^5 ≡ 4 * 9 ≡ 36 ≡ 10 (mod 13)
• 4^6 ≡ 4 * 10 ≡ 40 ≡ 1 (mod 13)

We found that 4^6 ≡ 1 (mod 13). This is a crucial observation because it reveals a cyclic pattern. In modular arithmetic, finding a power of the base that is congruent to 1 modulo m is a breakthrough. This is because any higher power can be broken down into multiples of this cycle. The power to which we raise 4 to get a remainder of 1 is known as the order of 4 modulo 13. This cyclic behavior simplifies the problem significantly, as we don't need to calculate 4^{103} directly. Instead, we can use the fact that 4^6 ≡ 1 (mod 13) to reduce the exponent 103. Modular arithmetic is a powerful tool in number theory, and this cyclic pattern is a prime example of its elegance and usefulness. Discovering these patterns is key to solving many problems involving large exponents and remainders.

The pattern repeats every 6 powers. Now we can use this to find the remainder of 4^{103} when divided by 13.

Calculating the Remainder of 4^{103}

Since we know that 4^6 ≡ 1 (mod 13), we can express 4^103} in terms of 4^6. To do this, divide 103 by 6 103 = 6 * 17 + 1. This means that 4^{103 can be written as (4⁶⁾{17} * 4^1. By breaking down the exponent into multiples of 6 and a remainder, we utilize the cyclic property we discovered earlier. The goal is to leverage the fact that 4^6 leaves a remainder of 1 when divided by 13. This allows us to significantly reduce the computational complexity. The process of dividing the exponent and using the quotient and remainder is a standard technique in modular arithmetic. This effectively allows us to simplify large exponents by exploiting the periodic behavior of powers modulo a given number. The remainder of this division directly corresponds to the power of 4 that we need to consider after accounting for the multiples of 6. This technique is crucial for efficiently solving remainder problems involving large exponents.

Therefore, 4^{103} = 4^(6*17 + 1) = (4⁶⁾{17} * 4^1.

Now we can use the property of modular arithmetic that (a * b) mod m = ((a mod m) * (b mod m)) mod m. So,

4^{103} ≡ (4⁶⁾{17} * 4^1 ≡ (1)^{17} * 4 ≡ 1 * 4 ≡ 4 (mod 13).

This result is crucial because it drastically simplifies our calculation. We've managed to find the remainder of a very large power of 4 by leveraging the cyclic pattern modulo 13. Now, instead of dealing with 4^{103}, we only need to consider the remainder 4. This significantly reduces the complexity of the problem. The elegance of modular arithmetic lies in its ability to transform complex calculations into simpler ones by focusing on remainders. The fact that (1)^{17} simplifies to 1 further streamlines the calculation. This step demonstrates the power of modular arithmetic in simplifying complex problems by reducing numbers to their remainders and exploiting periodic patterns. We are now well-positioned to complete the final calculation.

Final Calculation

We have found that 4^{103} ≡ 4 (mod 13) and 21 ≡ 8 (mod 13). Now we can find the remainder when 4^{103}(21) is divided by 13:

4^{103}(21) ≡ 4 * 8 ≡ 32 (mod 13).

Now we need to find the remainder when 32 is divided by 13. 32 divided by 13 gives a quotient of 2 and a remainder of 6.

Therefore, 32 ≡ 6 (mod 13).

So, the remainder when 4^103} + 4^{104} + 4^{105} is divided by 13 is 6. We've arrived at the final answer by combining the results of our modular arithmetic calculations. This final step synthesizes all the previous steps simplification, identification of cyclic patterns, and application of modular arithmetic properties. We multiplied the remainders of the individual factors, 4^{103 and 21, modulo 13, and then found the remainder of the product modulo 13. This systematic approach demonstrates the power of breaking down a complex problem into smaller, manageable steps. The remainder of 6 represents the final answer to the original question. This example illustrates the beauty of number theory and the power of modular arithmetic in solving problems involving large numbers and remainders. By following these logical steps, we successfully determined the remainder, providing a complete solution to the problem.

Conclusion

In conclusion, by simplifying the expression, utilizing modular arithmetic, and identifying the cyclic pattern of powers of 4 modulo 13, we found that the remainder when 4^{103} + 4^{104} + 4^{105} is divided by 13 is 6. This problem demonstrates the power and elegance of modular arithmetic in simplifying complex calculations involving exponents and remainders. The methodical approach of breaking down the problem into smaller, manageable steps is a key strategy for success in number theory and other mathematical domains. By understanding and applying these techniques, one can tackle a wide range of similar problems effectively. The ability to find patterns, simplify expressions, and apply fundamental mathematical principles is essential for problem-solving in mathematics. This exploration serves as a testament to the beauty and utility of mathematical tools in deciphering seemingly complex questions.