What is the remainder when 6599 is divided by 11 ?
10
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.
| Step | Calculation | Result (mod 11) |
|---|---|---|
| Simplify base | \(65 \div 11\) | \(10 \text{ or } -1\) |
| Substitute base | \(65^{99} \pmod{11}\) | \((-1)^{99} \pmod{11}\) |
| Evaluate power | \((-1)^{99}\) | \(-1\) |
| Positive remainder | \(-1 \pmod{11}\) | \(10\) |
Revision Table: Key Concepts
| Concept | Description | Example |
|---|---|---|
| Modular Arithmetic (\(a \equiv b \pmod{m}\)) | \(a\) and \(b\) have the same remainder when divided by \(m\). \(a-b\) is a multiple of \(m\). | \(15 \equiv 3 \pmod{12}\) because \(15 = 1 \times 12 + 3\). |
| Properties of Modulo |
|
|
| Negative Remainders | If \(a \equiv -b \pmod m\), then the positive remainder is \(m-b\). | \(-1 \pmod{11} \equiv -1 + 11 \pmod{11} \equiv 10 \pmod{11}\) |
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.
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