Understanding the Core Concepts

This problem involves testing divisibility of expressions. We will primarily use two key mathematical concepts:

1. Divisibility Property of Exponents: For any positive integers $a, b,$ and $n$, the expression $a^n - b^n$ is always divisible by $(a-b)$.
2. Binomial Theorem: For any non-negative integer $n$, the expansion of $(x+y)^n$ is given by: $(x+y)^n = \binom{n}{0}x^n y^0 + \binom{n}{1}x^{n-1}y^1 + \binom{n}{2}x^{n-2}y^2 + \dots + \binom{n}{n}x^0 y^n$ This theorem is particularly useful when checking divisibility by expressing one term as a sum involving the desired divisor (e.g., $13 = 1+12$ when checking divisibility by factors of $12$, like $144$).

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Analysis of Statement (S1): $2023^{2022} - 1999^{2022}$ is divisible by 8

Let's evaluate $S_1 = 2023^{2022} - 1999^{2022}$.

Method 1: Using the Divisibility Property of Exponents

1. Identify $a$ and $b$: We can let $a = 2023$ and $b = 1999$. The exponent $n = 2022$.
2. Calculate the difference $(a-b)$: $a - b = 2023 - 1999 = 24$
3. Apply the divisibility property: According to the property, $a^n - b^n$ is divisible by $a-b$. Therefore, $2023^{2022} - 1999^{2022}$ is divisible by $24$.
4. Check for divisibility by 8: Since $24$ is divisible by $8$ (as $24 = 3 \times 8$), any number that is a multiple of $24$ must also be a multiple of $8$. Thus, $S_1$ is divisible by $8$.

Method 2: Using Modular Arithmetic

1. Determine the remainders modulo 8: To check divisibility by 8, we can find the remainder of each base when divided by 8. For $2023$: $2023 = 8 \times 252 + 7$. So, $2023 \equiv 7 \pmod 8$. Alternatively, $2024$ is a multiple of $8$ ($2024 = 8 \times 253$). Thus, $2023 = 2024 - 1 \equiv -1 \pmod 8$. For $1999$: $1999 = 8 \times 249 + 7$. So, $1999 \equiv 7 \pmod 8$. Alternatively, $2000$ is a multiple of $8$ ($2000 = 8 \times 250$). Thus, $1999 = 2000 - 1 \equiv -1 \pmod 8$.
2. Substitute the modular equivalents into the expression: Since $2023 \equiv -1 \pmod 8$ and $1999 \equiv -1 \pmod 8$: $2023^{2022} - 1999^{2022} \equiv (-1)^{2022} - (-1)^{2022} \pmod 8$
3. Evaluate the powers: The exponent $2022$ is an even number. $(-1)^{2022} = 1$ So, the expression becomes: $\equiv 1 - 1 \pmod 8$ $\equiv 0 \pmod 8$
4. Conclusion: Since the remainder is $0$ when divided by $8$, Statement (S1) is correct.

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Analysis of Statement (S2): $13(13)^n - 12n - 13$ is divisible by 144 for infinitely many $n \in \mathbb{N}$

Let the expression be $E_n = 13^{n+1} - 12n - 13$. We need to check its divisibility by 144.

1. Rewrite $13^{n+1}$ using the Binomial Theorem: To check divisibility by 144, which is $12^2$, it's helpful to express $13$ as $(1+12)$. $13^{n+1} = (1+12)^{n+1}$ Now, we expand this using the Binomial Theorem $(a+b)^k = \binom{k}{0}a^k b^0 + \binom{k}{1}a^{k-1}b^1 + \binom{k}{2}a^{k-2}b^2 + \dots$: $13^{n+1} = \binom{n+1}{0} 1^{n+1} (12)^0 + \binom{n+1}{1} 1^n (12)^1 + \binom{n+1}{2} 1^{n-1} (12)^2 + \binom{n+1}{3} 1^{n-2} (12)^3 + \dots$ Let's simplify the first few terms: $13^{n+1} = 1 + (n+1) \cdot 12 + \frac{(n+1)n}{2} \cdot 144 + \frac{(n+1)n(n-1)}{6} \cdot 1728 + \dots$

2. Substitute the expansion back into $E_n$: $E_n = \left( 1 + 12(n+1) + \frac{n(n+1)}{2} \cdot 144 + \frac{n(n+1)(n-1)}{6} \cdot 1728 + \dots \right) - 12n - 13$

3. Simplify the expression: Expand $12(n+1)$ and group terms: $E_n = 1 + 12n + 12 + \frac{n(n+1)}{2} \cdot 144 + \frac{n(n+1)(n-1)}{6} \cdot 1728 + \dots - 12n - 13$ Combine the constant terms ($1+12-13$) and the terms involving $n$ ($12n - 12n$): $E_n = (1 + 12 - 13) + (12n - 12n) + \frac{n(n+1)}{2} \cdot 144 + \frac{n(n+1)(n-1)}{6} \cdot 1728 + \dots$ $E_n = 0 + 0 + \frac{n(n+1)}{2} \cdot 144 + \frac{n(n+1)(n-1)}{6} \cdot 1728 + \dots$ Notice that $1728 = 12 \times 144$. All terms from $\binom{n+1}{2}12^2$ onwards are multiples of $144$. Also, $\frac{n(n+1)}{2}$ is always an integer for any natural number $n$, as either $n$ or $n+1$ is even. Thus, we can factor out $144$ from the remaining terms: $E_n = 144 \left( \frac{n(n+1)}{2} + \frac{n(n+1)(n-1)}{6} \cdot 12 + \dots \right)$ The expression within the parenthesis is an integer for all $n \in \mathbb{N}$.

4. Conclusion: Since $E_n$ can be written as $144$ multiplied by an integer for all $n \in \mathbb{N}$, the expression is divisible by $144$ for all $n \in \mathbb{N}$. This clearly implies it is divisible by $144$ for infinitely many $n \in \mathbb{N}$. Therefore, Statement (S2) is correct.

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Final Answer

Based on our detailed analysis, both Statement (S1) and Statement (S2) are correct.

The correct option is (D).

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Tips and Common Mistakes

• For Statement S1:
  • Tip: For divisibility problems, especially with large numbers, always consider modular arithmetic first. It can significantly simplify calculations. The property $a^n - b^n$ divisible by $a-b$ is also a powerful tool.
  • Common Mistake: Overcomplicating by using binomial expansion when simpler methods suffice. The binomial expansion used in the provided solution does confirm divisibility by 24 (and thus 8), but modular arithmetic is more direct.
• For Statement S2:
  • Tip: When using the Binomial Theorem for divisibility, rewrite the base number such that one term in the binomial is the divisor or a factor of the divisor (e.g., $13 = 1+12$ for divisibility by 144). Expand enough terms to show the required divisibility ($12^2 = 144$).
  • Common Mistake: Not expanding enough terms. For a modulus like $M = k^2$, you typically need to expand terms up to $k^2$ to capture all relevant parts of the expression. Also, forgetting that binomial coefficients $\binom{N}{k}$ are always integers.

Summary and Key Takeaway

This problem highlights the versatility of number theory concepts, particularly modular arithmetic and the Binomial Theorem, in solving divisibility problems. By understanding the properties of exponents and judiciously applying binomial expansion, we can efficiently determine divisibility. Both statements presented in the question are mathematically sound, demonstrating divisibility as claimed.