Finding the Remainder Using Modular Arithmetic

The problem asks for the remainder when a large power, 65⁹⁹, is divided by 11. This type of problem can be efficiently solved using the concept of modular arithmetic.

Understanding Modular Arithmetic for Remainders

Modular arithmetic deals with remainders. The expression $a \equiv b \pmod{m}$ means that when $a$ is divided by $m$, the remainder is the same as when $b$ is divided by $m$. In other words, $a - b$ is divisible by $m$. We are looking for the value of $65^{99} \pmod{11}$.

Step 1: Simplify the Base Modulo 11

First, let's find the remainder of the base, 65, when divided by 11.

We divide 65 by 11:

$$65 = 11 \times 5 + 10$$

The remainder is 10. So, we can write:

$$65 \equiv 10 \pmod{11}$$

An equivalent way to express $10 \pmod{11}$ is $-1 \pmod{11}$, because $10 - (-1) = 11$, which is divisible by 11. Using $-1$ is often helpful for powers.

So, $65 \equiv 10 \equiv -1 \pmod{11}$.

Step 2: Evaluate the Power Modulo 11

Now we need to find the remainder of $65^{99}$ when divided by 11. Since $65 \equiv -1 \pmod{11}$, we can substitute:

$$65^{99} \equiv (-1)^{99} \pmod{11}$$

Step 3: Calculate the Power of -1

We need to calculate $(-1)^{99}$. When -1 is raised to an odd power, the result is -1.

$$(-1)^{99} = -1$$

Step 4: Express the Result as a Positive Remainder

So, we have:

$$65^{99} \equiv -1 \pmod{11}$$

Remainders in modular arithmetic are typically expressed as non-negative integers less than the modulus (in this case, less than 11). To convert -1 to a positive remainder modulo 11, we add the modulus:

$$-1 \equiv -1 + 11 \pmod{11}$$

$$-1 \equiv 10 \pmod{11}$$

Therefore, the remainder when $65^{99}$ is divided by 11 is 10.

Summary of the Calculation

• We want to find $65^{99} \pmod{11}$.
• Simplify the base: $65 \equiv 10 \pmod{11}$.
• Alternatively, $65 \equiv -1 \pmod{11}$.
• Substitute the equivalent base: $65^{99} \equiv (-1)^{99} \pmod{11}$.
• Calculate the power: $(-1)^{99} = -1$.
• Convert to a positive remainder: $-1 \equiv 10 \pmod{11}$.

The final remainder is 10.

Revision Table: Key Concepts

Additional Information: Other Methods for Remainders

While using $65 \equiv -1 \pmod{11}$ was the quickest method here because of the odd exponent, for other numbers and exponents, we might use Fermat's Little Theorem or Euler's Totient Theorem.

Fermat's Little Theorem: If $p$ is a prime number, then for any integer $a$ not divisible by $p$, $a^{p-1} \equiv 1 \pmod{p}$.

In our problem, $p=11$ (which is prime) and $a=65$ (not divisible by 11). So, $65^{11-1} \equiv 65^{10} \equiv 1 \pmod{11}$.

We could write $65^{99} = 65^{10 \times 9 + 9} = (65^{10})^9 \times 65^9$.

Using modular properties:

$$65^{99} \equiv (65^{10})^9 \times 65^9 \pmod{11}$$

Since $65^{10} \equiv 1 \pmod{11}$, we get:

$$65^{99} \equiv (1)^9 \times 65^9 \pmod{11}$$

$$65^{99} \equiv 1 \times 65^9 \pmod{11}$$

$$65^{99} \equiv 65^9 \pmod{11}$$

Now, we need to find $65^9 \pmod{11}$. We know $65 \equiv 10 \pmod{11}$, so:

$$65^9 \equiv 10^9 \pmod{11}$$

Since $10 \equiv -1 \pmod{11}$, we have $10^9 \equiv (-1)^9 \equiv -1 \pmod{11}$.

And $-1 \equiv 10 \pmod{11}$.

Both methods lead to the same remainder, 10.