Calculating the Remainder of 35×57×79 ÷ 11 Using Modular Arithmetic
Introduction
In mathematics, particularly in number theory and modular arithmetic, finding the remainder of a large number when divided by a smaller number is a common yet challenging task. This article will walk you through the process of calculating 35×57×79R when divided by 11, employing the principles of modular arithmetic. We will break down the problem into smaller, manageable parts and apply modular exponentiation to simplify the calculation.
Step-by-Step Calculation
First, let's start by breaking down the expression into its components and simplifying them one by one using modular arithmetic rules.
Modular Exponentiation of 35 ≡ 1 (mod 11)
31equiv 3mod{11}
32equiv 9mod{11}
33equiv 3times9equiv 5mod{11}
34equiv 3times5equiv 4mod{11}
35equiv 3times4equiv 1mod{11}
Modular Exponentiation of 57 ≡ 3 (mod 11)
51equiv 5mod{11}
52equiv 3mod{11}
53equiv 5times3equiv 4mod{11}
54equiv 5times4equiv 9mod{11}
55equiv 5times9equiv 1mod{11}
57equiv msup>51equiv 3mod{11}
Modular Exponentiation of 79 ≡ 8 (mod 11)
71equiv 7mod{11}
72equiv 5mod{11}
73equiv 7times5equiv 2mod{11}
74equiv 7times2equiv 3mod{11}
75equiv 7times3equiv 10mod{11}
76equiv 7times10equiv 4mod{11}
77equiv 7times4equiv 6mod{11}
78equiv 7times6equiv 9mod{11}
79equiv 7times9equiv 8mod{11}
Final Step: Combining the Results
Now we combine the results of the modular exponentiations:
R 3^5 times 5^7 times 7^9 mod{11} 3^5 mod{11} times 5^7 mod{11} times 7^9 mod{11} 1 times 3 times 8 mod{11} 24 mod{11} 2
Conclusion
The remainder when 3^5 times 5^7 times 7^9 is divided by 11 is 2. This method highlights how modular arithmetic can simplify complex calculations and make them more manageable.