Key Concept: Binomial Theorem and Modular Arithmetic

This problem involves finding the remainder of an algebraic expression when divided by a specific number (9). The core concept used here is a powerful application of the Binomial Theorem in conjunction with Modular Arithmetic.

The Binomial Theorem allows us to expand expressions of the form $(a+b)^n$. $(a+b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \dots + \binom{n}{n} a^0 b^n$ When we need to find the remainder upon division by a number, say $M$, we can strategically rewrite the terms in the form $(kM \pm x)^N$. In such an expansion, every term containing $kM$ (or its powers) as a factor will be a multiple of $M$, and thus will contribute 0 to the remainder. The remainder will then depend solely on the terms that do not contain $kM$. Specifically, a crucial property for finding remainders is: If $X$ is a multiple of $M$, then $(X \pm k)^N \equiv (\pm k)^N \pmod M$. This means we only need to consider the remainder of the $k^N$ part.

Step-by-Step Solution

We need to find the remainder when ${8^{2n}} - {\left( {62} \right)^{2n + 1}}$ is divided by 9. We will analyze each term separately using the Binomial Theorem.

Step 1: Rewrite the base numbers as a multiple of 9 plus or minus 1.

• For the first term, $8^{2n}$: We can express $8$ in relation to $9$ as $(9 - 1)$. Therefore, the first term becomes ${8^{2n}} = {(9 - 1)^{2n}}$.
• For the second term, ${\left( {62} \right)^{2n + 1}}$: We need to relate $62$ to $9$. $62$ is close to $63$, which is a multiple of $9$ ($63 = 9 \times 7$). So, we can express $62$ as $(63 - 1)$. Therefore, the second term becomes ${\left( {62} \right)^{2n + 1}} = {\left( {63 - 1} \right)^{2n + 1}}$.

Step 2: Apply the Binomial Theorem to each term to find its remainder when divided by 9.

• Analyzing the first term: ${(9 - 1)^{2n}}$ Using the Binomial Theorem for $(a-b)^N = \sum_{k=0}^N \binom{N}{k} a^{N-k} (-b)^k$ (where $a=9$, $b=1$, and $N=2n$): ${(9 - 1)^{2n} = \binom{2n}{0} (9)^{2n} (-1)^0 + \binom{2n}{1} (9)^{2n-1} (-1)^1 + \dots + \binom{2n}{2n-1} (9)^1 (-1)^{2n-1} + \binom{2n}{2n} (9)^0 (-1)^{2n}}$ When this entire expansion is divided by 9: Every term from the first term $\binom{2n}{0} (9)^{2n} (-1)^0$ down to the second-to-last term $\binom{2n}{2n-1} (9)^1 (-1)^{2n-1}$ contains a factor of 9 (or a higher power of 9). This means all these terms are perfectly divisible by 9, contributing a remainder of 0. The remainder will thus be determined by the last term, which does not contain a factor of 9: ${(9 - 1)^{2n} \equiv \binom{2n}{2n} (9)^0 (-1)^{2n} \pmod 9}$ Since $\binom{2n}{2n} = 1$, $(9)^0 = 1$, and $2n$ is an even integer, ${(-1)^{2n} = 1}$: ${(9 - 1)^{2n} \equiv 1 \times 1 \times 1 \pmod 9}$ ${(9 - 1)^{2n} \equiv 1 \pmod 9}$ So, the remainder of ${8^{2n}}$ when divided by 9 is 1.

• Analyzing the second term: ${\left( {63 - 1} \right)^{2n + 1}}$ Using the Binomial Theorem (where $a=63$, $b=1$, and $N=2n+1$): {\left( {63 - 1} \right)^{2n + 1}} = \binom{2n+1}{0} (63)^{2n+1} (-1)^0 + \binom{2n+1}{1} (63)^{2n} (-1)^1 + \dots + \binom{2n+1}{2n} (63)^1 (-1)^{2n} + \binom{2n+1}{2n+1} (63)^0 (-1)^{2n+1}} When this entire expansion is divided by 9: Since $63$ is a multiple of $9$ ($63 = 7 \times 9$), every term from the first term $\binom{2n+1}{0} (63)^{2n+1} (-1)^0$ down to the second-to-last term $\binom{2n+1}{2n} (63)^1 (-1)^{2n}$ contains a factor of 63 (or a higher power), and thus is a multiple of 9. These terms contribute 0 to the remainder. The remainder will be determined by the last term: ${\left( {63 - 1} \right)^{2n + 1}} \equiv \binom{2n+1}{2n+1} (63)^0 (-1)^{2n+1} \pmod 9$ We know that $\binom{2n+1}{2n+1} = 1$, $(63)^0 = 1$, and $2n+1$ is an odd integer, so ${(-1)^{2n + 1} = -1}$: ${\left( {63 - 1} \right)^{2n + 1}} \equiv 1 \times 1 \times (-1) \pmod 9$ ${\left( {63 - 1} \right)^{2n + 1}} \equiv -1 \pmod 9$ So, the remainder of ${\left( {62} \right)^{2n + 1}}$ when divided by 9 is -1 (which is equivalent to 8, as $-1 + 9 = 8$).

Step 3: Combine the remainders for the original expression.

Now we substitute the individual remainders back into the original expression: ${8^{2n}} - {\left( {62} \right)^{2n + 1}}$ $\equiv 1 - (-1) \pmod 9$ $\equiv 1 + 1 \pmod 9$ $\equiv 2 \pmod 9$

Thus, the remainder when ${8^{2n}} - {\left( {62} \right)^{2n + 1}}$ is divided by 9 is 2.

Alternative Approach: Direct Modular Arithmetic

A more direct and often quicker approach for remainder problems is to use properties of modular arithmetic directly:

1. Simplify bases modulo 9:

   • $8 \equiv -1 \pmod 9$
   • $62 = 6 \times 9 + 8 \equiv 8 \equiv -1 \pmod 9$

2. Apply powers:

   • $8^{2n} \equiv (-1)^{2n} \pmod 9$ Since $2n$ is always an even integer, $(-1)^{2n} = 1$. So, $8^{2n} \equiv 1 \pmod 9$.
   • ${\left( {62} \right)^{2n + 1}} \equiv (-1)^{2n + 1} \pmod 9$ Since $2n+1$ is always an odd integer, $(-1)^{2n + 1} = -1$. So, ${\left( {62} \right)^{2n + 1}} \equiv -1 \pmod 9$.

3. Combine: ${8^{2n}} - {\left( {62} \right)^{2n + 1}} \equiv 1 - (-1) \pmod 9$ $\equiv 1 + 1 \pmod 9$ $\equiv 2 \pmod 9$ Both methods yield the same correct remainder.

Tips for Success & Common Pitfalls

• Choose the right form: When using the Binomial Theorem for remainders, try to express the base as $(M \pm 1)$ or $(M \pm k)$, where $M$ is a multiple of the divisor. This simplifies the expansion significantly.
• Negative Remainders: A remainder like -1 is mathematically valid. To convert it to a positive remainder (which is typically expected in answers), add the divisor: $-1 + 9 = 8$. However, for intermediate steps, negative remainders can simplify calculations, as seen in the modular arithmetic approach.
• Parity of Exponents: Always pay close attention to whether the exponent is even or odd when dealing with $(-1)^N$.
• Binomial Expansion vs. Modular Arithmetic: While the Binomial Theorem provides the fundamental proof, direct modular arithmetic ($a \equiv b \pmod m$) is often much faster and less prone to calculation errors for remainder problems.

Summary & Key Takeaway

By strategically rewriting the bases of the powers as "multiple of 9 plus or minus 1" and then applying the Binomial Theorem (or direct modular arithmetic), we efficiently determined that all terms except the final $\pm 1$ cancel out modulo 9. The expression simplifies to $1 - (-1) = 2 \pmod 9$. This demonstrates the power of simplifying numbers modulo the divisor before performing operations, especially with large powers.