Finding the Remainder of Complex Modular Arithmetic Problems

In this article, we will explore how to solve complex modular arithmetic problems using modular arithmetic. Specifically, we will focus on finding the remainder when large powers of numbers are divided by other numbers. We will break down the processes step-by-step using examples and a detailed approach to make the calculations clear and understandable.

Example 1: Finding the Remainder When ( 1234^{5678} ) is Divided by 9

Let's start by finding the remainder when ( 1234^{5678} ) is divided by 9.

First, we reduce 1234 modulo 9:

To find ( 1234 mod 9 ), sum the digits: ( 1 2 3 4 10 ). Since ( 10 equiv 1 mod 9 ), we have ( 1234 equiv 1 mod 9 ).

Next, we simplify ( 1234^{5678} mod 9 ):

Given ( 1234 equiv 1 mod 9 ), we have ( 1234^{5678} equiv 1^{5678} mod 9 ). Since any number raised to any power is still 1 modulo 9 if the base is 1, we get ( 1234^{5678} equiv 1 mod 9 ).

Therefore, the remainder when ( 1234^{5678} ) is divided by 9 is 1.

Example 2: Finding the Remainder When ( 1234 times 5678^9 ) is Divided by 2019

To find the remainder when ( 1234 times 5678^9 ) is divided by 2019, we will use modular arithmetic to break down the problem into smaller parts and compute the result step-by-step.

First, we reduce ( 1234^2 ) modulo 2019:

[ 1234^2 equiv 430 mod 2019 ]

Next, we reduce ( 5678^2 ) modulo 2019:

[ 5678^2 equiv 292 mod 2019 ]

Then, we use these values to find ( 1234^9 times 5678^9 mod 2019 ):

[ 1234^9 equiv 1799 mod 2019 ]

[ 5678^9 equiv 1799 mod 2019 ]

[ 1234^9 times 5678^9 equiv 1799 times 1799 mod 2019 equiv 1799 mod 2019 ]

Thus, the remainder when ( 1234 times 5678^9 ) is divided by 2019 is 1799.

Example 3: Finding the Remainder of a Complex Expression

We will now explore a more complex expression ( 1234 times 5678 equiv -785 times -379 mod 2019 ):

First, reduce the product modulo 2019:

[ -785 times -379 equiv 361015 mod 2019 equiv 1799 mod 2019 ]

Then, use the same approach for the expression ( 1234^9 times 5678^9 ):

[ 1234^9 equiv 325312 equiv 360 mod 2019 ]

[ 5678^9 equiv 1190218 equiv 392 mod 2019 ]

[ 1234^9 times 5678^9 equiv 360 times 392 equiv 141120 equiv 1799 mod 2019 ]

In conclusion, the remainder when ( 1234 times 5678^9 ) is divided by 2019 is 1799.

Conclusion

By breaking down complex modular arithmetic problems into simpler parts, we can find the remainders efficiently. The key steps include reducing the base modulo the given number, applying modular exponentiation, and using modular properties to simplify large calculations.