Calculating Remainder Using Modular Arithmetic and Properties

We are asked to find the remainder when the expression $({17}^{29}+{19}^{29})$ is divided by 18. This is a problem that can be efficiently solved using the principles of modular arithmetic.

Method 1: Using Modular Arithmetic Properties

The core idea of modular arithmetic is looking at the remainder when a number is divided by another number. We want to find $({17}^{29}+{19}^{29})$ modulo 18.

• Let's consider the base 17 with respect to the divisor 18. $$17\equiv -118$$ This is because 17 is one less than 18.
• Now, let's raise this congruence to the power of 29: $${17}^{29}\equiv {(-1)}^{29}18$$
• Since 29 is an odd exponent, ${(-1)}^{29}$ is equal to -1. $${17}^{29}\equiv -118$$
• Next, consider the base 19 with respect to the divisor 18. $$19\equiv 118$$ This is because 19 is one more than 18.
• Now, raise this congruence to the power of 29: $${19}^{29}\equiv 1^{29}18$$
• Since any power of 1 is 1, $1^{29}$ is equal to 1. $${19}^{29}\equiv 118$$
• Now, we need to find the remainder of the sum $({17}^{29}+{19}^{29})$. We can add the congruences: $${17}^{29}+{19}^{29}\equiv (-1)+118$$
• Simplifying the right side: $${17}^{29}+{19}^{29}\equiv 018$$

This means that when $({17}^{29}+{19}^{29})$ is divided by 18, the remainder is 0.

Method 2: Using Algebraic Identity

There is a useful algebraic identity that states that for any positive odd integer $n$, the expression $a^n+b^n$ is always divisible by $(a+b)$.

• In our problem, we have the expression ${17}^{29}+{19}^{29}$.
• Here, $a=17$, $b=19$, and the exponent $n=29$.
• Since 29 is an odd number, we can apply the identity.
• According to the identity, ${17}^{29}+{19}^{29}$ is divisible by $(a+b)$, which is $(17+19)$.
• Let's calculate the sum $(17+19)$: $$17+19=36$$
• So, ${17}^{29}+{19}^{29}$ is divisible by 36.
• Now, consider the divisor in the question, which is 18. Since 36 is a multiple of 18 ($36=2\times 18$), any number that is divisible by 36 is also divisible by 18.
• Therefore, ${17}^{29}+{19}^{29}$ is divisible by 18.

When a number is divisible by another number, the remainder is always 0.

Both methods confirm that the remainder when $({17}^{29}+{19}^{29})$ is divided by 18 is 0.

Revision Table: Key Concepts for Remainder Problems

Additional Information on Remainder Calculations

Calculating remainders, especially with large numbers and exponents, is a common topic in number theory and is often tested in exams. Here are some related concepts:

• Fermat's Little Theorem: If $p$ is a prime number, then for any integer $a$ not divisible by $p$, we have $a^{p-1}\equiv 1pmod><mi>p</mi>$. This can simplify calculations with prime moduli.
• Euler's Totient Theorem: This is a generalization of Fermat's Little Theorem. If $n$ is a positive integer, then for any integer $a$ coprime to $n$, we have $a^{\phi \left(n\right)}\equiv 1pmod><mi>n</mi>$, where $\phi \left(n\right)$ is Euler's totient function. This helps with composite moduli like 18.
• Chinese Remainder Theorem: Useful for finding a number that has specific remainders when divided by several different numbers.
• Negative Remainders: As seen in the solution, sometimes using negative remainders (like -1 mod 18) can simplify calculations significantly. The final positive remainder is obtained by adding the modulus if the result is negative (e.g., $-1\equiv -1+18\equiv 17pmod><mn>18</mn>$). In this problem, the sum was 0 mod 18, which is a standard remainder.