What is the remainder when 27 27 - 15 27 is divided by 6?
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The question asks for the remainder when a mathematical expression, specifically \(27^{27} - 15^{27}\), is divided by 6. This type of problem is best solved using the principles of modular arithmetic. Modular arithmetic helps us work with remainders efficiently, especially with large numbers and exponents.
To find the remainder of the entire expression when divided by 6, we first need to consider the remainders of the bases, 27 and 15, when they are divided by 6.
A key property of modular arithmetic states that if two numbers are congruent modulo a number \(m\), then their positive integer powers are also congruent modulo \(m\). Mathematically, if \(a \equiv b \pmod{m}\), then \(a^n \equiv b^n \pmod{m}\) for any positive integer \(n\).
In our case, we have:
Now we need to find the remainder of the difference, \(27^{27} - 15^{27}\), when divided by 6. Using the congruences we found:
\(27^{27} - 15^{27} \equiv (3^{27}) - (3^{27}) \pmod{6}\)
The expression simplifies to:
\(27^{27} - 15^{27} \equiv 0 \pmod{6}\)
This means that the remainder when \(27^{27} - 15^{27}\) is divided by 6 is 0.
Let's summarize the process:
| Step | Calculation | Result Modulo 6 |
|---|---|---|
| Base 27 | \(27 \pmod{6}\) | 3 |
| Base 15 | \(15 \pmod{6}\) | 3 |
| Power 27 (\(27^{27}\)) | \(27^{27} \equiv 3^{27} \pmod{6}\) | \(3^{27} \pmod{6}\) |
| Power 27 (\(15^{27}\)) | \(15^{27} \equiv 3^{27} \pmod{6}\) | \(3^{27} \pmod{6}\) |
| Difference | \(27^{27} - 15^{27} \equiv 3^{27} - 3^{27} \pmod{6}\) | \(0 \pmod{6}\) |
The final remainder is 0.
| Concept | Explanation | Example |
|---|---|---|
| Modular Arithmetic | A system of arithmetic for integers, where numbers "wrap around" upon reaching a certain value (the modulus). It's focused on remainders. | \(17 \equiv 5 \pmod{12}\) (17 hours after 12 o'clock is 5 o'clock) |
| Congruence | Two integers \(a\) and \(b\) are congruent modulo \(m\) if their difference \(a-b\) is divisible by \(m\), or if they have the same remainder when divided by \(m\). | \(27 \equiv 15 \pmod{6}\) because \(27-15 = 12\), which is divisible by 6. Also, \(27 \div 6\) has remainder 3, and \(15 \div 6\) has remainder 3. |
| Congruence of Powers | If \(a \equiv b \pmod{m}\), then \(a^n \equiv b^n \pmod{m}\) for any positive integer \(n\). | Since \(2 \equiv 5 \pmod{3}\), then \(2^2 \equiv 5^2 \pmod{3}\) (\(4 \equiv 25 \pmod{3}\), \(1 \equiv 1 \pmod{3}\)). |
Modular arithmetic is a fundamental tool in number theory and has applications in cryptography, computer science, and other fields. It simplifies calculations involving large numbers by focusing only on their remainders.
Some basic properties include:
In our problem, we used the subtraction property after applying the power property to simplify \(27^{27} - 15^{27}\) modulo 6.
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select the correct answer using the code given below: