Key Concept: Binomial Theorem for Remainders

To find the remainder when a large power of a number is divided by another number, we often use the Binomial Theorem. The core idea is to express the base of the power as $(M \pm k)$, where $M$ is a multiple of the divisor and $k$ is a small integer. The Binomial Theorem states that for any positive integer $n$: $(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n-1}ab^{n-1} + b^n$ When we divide $(M \pm k)^n$ by $M$, all terms in the expansion except the last one, $(\pm k)^n$, will contain $M$ as a factor, and thus will be perfectly divisible by $M$. Therefore, the remainder will be determined solely by the remainder of $(\pm k)^n$ when divided by $M$. This can be written using modular arithmetic as: $(M \pm k)^n \equiv (\pm k)^n \pmod M$

Step-by-Step Solution

Given the expression: ${{{{\left( {27} \right)}^{999}}} \over 7}$ We need to find the remainder when $(27)^{999}$ is divided by $7$.

Step 1: Express the base in terms of the divisor The divisor is $7$. We look for a multiple of $7$ that is close to $27$. We know that $7 \times 4 = 28$. So, we can rewrite $27$ as $28 - 1$. This step is crucial because $28$ is a multiple of $7$, which will simplify the application of the Binomial Theorem. ${\frac{{{{\left( {27} \right)}^{999}}}}{7}} = {\frac{{{{\left( {28 - 1} \right)}^{999}}}}{7}}$

Step 2: Apply the Binomial Theorem Now, we expand $(28 - 1)^{999}$ using the Binomial Theorem. $(28 - 1)^{999} = \binom{999}{0} (28)^{999} (-1)^0 + \binom{999}{1} (28)^{998} (-1)^1 + \dots + \binom{999}{998} (28)^1 (-1)^{998} + \binom{999}{999} (28)^0 (-1)^{999}$ Notice that every term in this expansion, except for the very last term $\binom{999}{999} (28)^0 (-1)^{999}$, contains a factor of $28$. Since $28$ is a multiple of $7$, all these terms are perfectly divisible by $7$. Let $K$ be an integer representing the sum of all terms containing $28$ as a factor. Then, $K$ is a multiple of $28$, and thus also a multiple of $7$. So, we can write: $(28 - 1)^{999} = (\text{a multiple of } 28) + \binom{999}{999} (28)^0 (-1)^{999}$ $(28 - 1)^{999} = 28\lambda + 1 \cdot 1 \cdot (-1)^{999}$ Where $28\lambda$ represents the sum of all terms divisible by $28$. Since $999$ is an odd number, $(-1)^{999} = -1$. Therefore, $(28 - 1)^{999} = 28\lambda - 1$

Step 3: Determine the remainder Now we need to find the remainder when $28\lambda - 1$ is divided by $7$. Since $28\lambda$ is a multiple of $7$, it leaves a remainder of $0$ when divided by $7$. So, the remainder of $(28\lambda - 1)$ when divided by $7$ is the same as the remainder of $(-1)$ when divided by $7$. {{\left( {28 - 1} \right)^{999}}} \over 7} = {{28\,\lambda - 1} \over 7} A remainder cannot be negative in the conventional sense. A negative remainder like $-1$ means that the number is $1$ less than a multiple of the divisor. To find the positive remainder, we add the divisor to the negative remainder until we get a positive value. In this case, $-1 \pmod 7$: $-1 + 7 = 6$ Alternatively, we can express $28\lambda - 1$ as: $28\lambda - 1 = 28\lambda - 7 + 6$ $28\lambda - 1 = 7(4\lambda - 1) + 6$ When this expression is divided by $7$, the term $7(4\lambda - 1)$ is perfectly divisible by $7$, leaving a remainder of $0$. Thus, the remainder is $6$. $\therefore\,\,\,$ Remainder $= 6$

Oops, I noticed a discrepancy. The correct answer was A (1), but my derived remainder is 6. Let me re-evaluate the calculation carefully.

If the remainder is $6$, then $(27)^{999} \equiv 6 \pmod 7$. However, the provided Correct Answer is A (1). Let me re-check the logic. $(27)^{999} \pmod 7$ $27 \equiv -1 \pmod 7$ So, $(27)^{999} \equiv (-1)^{999} \pmod 7$ Since $999$ is an odd number, $(-1)^{999} = -1$. So, $(27)^{999} \equiv -1 \pmod 7$. A remainder of $-1$ is equivalent to $6 \pmod 7$. So, the remainder is $6$.

This contradicts the given correct answer (A) 1. Let me carefully inspect the problem statement and the provided solution again. The provided solution also derives 6. Current Solution: We have, ${{{{\left( {27} \right)}^{999}}} \over 7}$ = ${{{{\left( {28 - 1} \right)}^{999}}} \over 7}$ = ${{28\,\lambda - 1} \over 7}$ = ${{28\,\lambda - 7 + 7 - 1} \over \lambda }$ = ${{7\left( {4\lambda - 1} \right) + 6} \over 7}$ $\therefore\,\,\,$ Remainder = 6

Both my derivation and the provided "Current Solution" arrive at 6. If the problem states "Correct Answer: A" and Option A is "1", then there is a mismatch between the problem's expected answer and the solution derived. I must assume the problem's stated "Correct Answer: A" (which means 1) is intended to be the actual correct answer, and the provided "Current Solution" is incorrect or there's a misunderstanding.

Let's re-evaluate the problem assuming the answer IS 1. If the remainder is 1, then: $(27)^{999} \equiv 1 \pmod 7$. This would mean that $27 \equiv 1 \pmod 7$. But $27 = 3 \times 7 + 6$, so $27 \equiv 6 \pmod 7$ or $27 \equiv -1 \pmod 7$. So, $(27)^{999} \equiv (-1)^{999} \equiv -1 \pmod 7 \equiv 6 \pmod 7$.

There seems to be a fundamental conflict between the question's stated "Correct Answer: A" (which corresponds to 1) and the actual mathematical calculation. If the answer were 1, then the problem might have been something like $(29)^{999} \pmod 7$ since $29 \equiv 1 \pmod 7$. Or, if the power was even, e.g., $(27)^{998} \pmod 7 \equiv (-1)^{998} \pmod 7 \equiv 1 \pmod 7$.

Given that the request is to rewrite the solution, and the current solution arrives at 6, it's highly likely that the "Correct Answer: A" and "1" is a mistake in the prompt's Correct Answer field, and the solution which results in 6 is what is expected. I should follow the provided "Current Solution" logic, which gives 6. And the request is to elaborate on that solution.

Let me proceed with the explanation that the remainder is 6, as derived by the provided solution and my own re-calculation. The "Correct Answer: A" might be a metadata error. I will explicitly state the derivation of 6.

Let's check for any mistakes in my understanding of the binomial theorem application. $(x-1)^n = \sum_{k=0}^n \binom{n}{k} x^{n-k} (-1)^k$ All terms for $k < n$ have a factor of $x$. The last term is $\binom{n}{n} x^{n-n} (-1)^n = 1 \cdot x^0 \cdot (-1)^n = (-1)^n$. So, $(x-1)^n \equiv (-1)^n \pmod x$. Here $x=28$, $n=999$. $(28-1)^{999} \equiv (-1)^{999} \pmod {28}$. So $(27)^{999} \equiv -1 \pmod {28}$. If $(27)^{999} \equiv -1 \pmod {28}$, then it also means $(27)^{999} \equiv -1 \pmod 7$ (since 7 divides 28). And $-1 \pmod 7 \equiv 6 \pmod 7$. So, the remainder is indeed 6.

I will proceed with explaining the solution that yields 6. I will add a tip about checking the parity of the exponent.

Corrected thought process: The user provided:

• Question: $(27)^{999}$ divided by 7
• Options: (A) 1, (B) 2, (C) 3, (D) 6
• Correct Answer: A (This means the user thinks the correct answer is 1, but the "Current Solution" yields 6)
• Current Solution: Shows remainder as 6.

My task is to rewrite the given solution (which leads to 6) to be more elaborate, clear, and educational. I should adhere to the provided current solution's logic, which leads to a remainder of 6, despite the "Correct Answer: A" pointing to 1. It's likely the Correct Answer field is mislabeled in the prompt. I will explain the solution that derives 6. I will not try to derive 1, as that would contradict the provided "Current Solution" and the mathematical facts.

Let me re-confirm all parts of the plan given this slight nuance.

1. Identify Key Concept: Binomial Theorem for Remainders. (Done)
2. Step-by-step Working with Explanations: (Done)
   • Step 1: Rewrite 27 as (28 - 1). Explain why this is useful (28 is a multiple of 7). (Done)
   • Step 2: Apply Binomial Theorem to $(28-1)^{999}$. Explain how terms involving 28 become divisible by 7, and why $(-1)^{999}$ is $-1$. (Done)
   • Step 3: Simplify to find the remainder. Explain how to handle negative remainders ($-1 \pmod 7 \equiv 6 \pmod 7$). (Done)
3. LaTeX Formatting: (Will ensure this during writing)
4. Tips/Common Mistakes: Add a section for common pitfalls or useful tips related to remainder problems, especially concerning negative remainders and the parity of the exponent. (Will add)
5. Summary/Key Takeaway: Conclude with a brief summary of the method. (Will add)
6. Format with clear section headers (bold markdown): (Will ensure this during writing)
7. Output only the rewritten solution: (Will ensure this)

Okay, I'm confident in explaining the solution that yields 6.Key Concept: Binomial Theorem for Remainders

To efficiently determine the remainder when a large power of an integer is divided by another integer, we can leverage the Binomial Theorem. The fundamental strategy involves expressing the base of the power in the form $(M \pm k)$, where $M$ is a multiple of the divisor and $k$ is a relatively small integer.

The Binomial Theorem states that for any positive integer $n$: $(a+b)^n = \binom{n}{0}a^n b^0 + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n-1}a^1 b^{n-1} + \binom{n}{n}a^0 b^n$ When we apply this to $(M \pm k)^n$ and divide the result by $M$, all terms in the expansion that contain $M$ as a factor (i.e., every term except possibly the last one, $(\pm k)^n$) will yield a remainder of $0$. Therefore, the overall remainder is solely determined by the remainder of $(\pm k)^n$ when divided by $M$. This concept is concisely expressed using modular arithmetic as: $(M \pm k)^n \equiv (\pm k)^n \pmod M$

Step-by-Step Solution

We are asked to find the remainder when $(27)^{999}$ is divided by $7$. ${{{{\left( {27} \right)}^{999}}} \over 7}$

Step 1: Express the Base in terms of the Divisor Our divisor is $7$. We need to find a multiple of $7$ that is close to the base, $27$. We know that $7 \times 4 = 28$. Thus, we can rewrite $27$ as $28 - 1$. This transformation is strategic because $28$ is a direct multiple of our divisor $7$, which will simplify the application of the Binomial Theorem. ${\frac{{{{\left( {27} \right)}^{999}}}}{7}} = {\frac{{{{\left( {28 - 1} \right)}^{999}}}}{7}}$

Step 2: Apply the Binomial Theorem Now, we expand the numerator, $(28 - 1)^{999}$, using the Binomial Theorem where $a=28$, $b=-1$, and $n=999$. $(28 - 1)^{999} = \binom{999}{0} (28)^{999} (-1)^0 + \binom{999}{1} (28)^{998} (-1)^1 + \dots + \binom{999}{998} (28)^1 (-1)^{998} + \binom{999}{999} (28)^0 (-1)^{999}$ Observe that every term in this expansion, except the very last term $\binom{999}{999} (28)^0 (-1)^{999}$, contains $28$ as a factor. Since $28$ is a multiple of $7$, all these terms are perfectly divisible by $7$ and contribute a remainder of $0$. Let's denote the sum of all terms containing $28$ as a factor by $28\lambda$, where $\lambda$ is some integer. So, $28\lambda$ is clearly divisible by $7$. The expression simplifies to: $(28 - 1)^{999} = 28\lambda + \binom{999}{999} (28)^0 (-1)^{999}$ Recall that $\binom{n}{n} = 1$ and any non-zero number raised to the power of $0$ is $1$. So, $\binom{999}{999} = 1$ and $(28)^0 = 1$. Now we evaluate $(-1)^{999}$. Since $999$ is an odd number, $(-1)^{999} = -1$. Substituting these values: $(28 - 1)^{999} = 28\lambda + 1 \cdot 1 \cdot (-1)$ $(28 - 1)^{999} = 28\lambda - 1$

Step 3: Determine the Final Remainder We now need to find the remainder when $28\lambda - 1$ is divided by $7$. Since $28\lambda$ is a multiple of $7$, it leaves a remainder of $0$ when divided by $7$. Therefore, the remainder of the entire expression is determined by the remainder of $-1$ when divided by $7$. ${{{{\left( {28 - 1} \right)}^{999}}} \over 7} = {{28\,\lambda - 1} \over 7}$ A conventional remainder must be a non-negative integer and less than the divisor. A negative remainder like $-1$ indicates that the number is $1$ less than a multiple of the divisor. To find the equivalent positive remainder, we add the divisor to the negative remainder. So, $-1 \pmod 7$ is equivalent to $-1 + 7 = 6 \pmod 7$. Alternatively, we can express $28\lambda - 1$ in the form $7q + r$ where $r \ge 0$: $28\lambda - 1 = (28\lambda - 7) + 6$ $28\lambda - 1 = 7(4\lambda - 1) + 6$ Here, $7(4\lambda - 1)$ is a multiple of $7$. When this expression is divided by $7$, the term $7(4\lambda - 1)$ is perfectly divisible, leaving a remainder of $0$. The remaining term is $6$. Thus, the remainder is $6$. $\therefore\,\,\,$ Remainder $= 6$

Tips and Common Mistakes

• Choose M wisely: Always try to choose $M$ such that $M$ is a multiple of the divisor and $(\pm k)$ is a small integer, preferably $1$ or $-1$, to simplify calculations.
• Handle negative remainders: If your calculation yields a negative remainder (e.g., $-1$), always convert it to a positive remainder by adding the divisor. For example, $-1 \pmod 7 \equiv 6 \pmod 7$, and $-2 \pmod 5 \equiv 3 \pmod 5$.
• Parity of the exponent: Pay close attention to whether the exponent $n$ is even or odd when you have $(-1)^n$. If $n$ is even, $(-1)^n = 1$. If $n$ is odd, $(-1)^n = -1$. This is a common source of errors.

Summary/Key Takeaway

This problem demonstrates an effective application of the Binomial Theorem combined with modular arithmetic principles to find remainders for large powers. By expressing the base as $(M \pm k)$ where $M$ is a multiple of the divisor, the problem simplifies significantly, reducing the computation to finding the remainder of $(\pm k)^n$. Remember to correctly handle negative remainders by converting them to their positive equivalents.