Question

Statement 1 : \(2^{513}+2^{013}+8^{13}+3^{13}\) is divisible by \(7\).
Statement 2 : The value of integral part of \((7+4\sqrt{3})^{25}\) is an odd number.

• A. Both Statements are correct
• B. Both Statements are false
• C. Statement 1 is false and Statement 2 is correct
• D. Statement 1 is correct and Statement 2 is false

Hint:

For divisibility problems, always reduce powers using {modular cycles}. For expressions like \((a+\sqrt{b})^n\), use the {conjugate pair trick}.

Answer 1

The Correct Option is A

To determine whether both given statements are correct, we will analyze each statement individually:

Analysis of Statement 1: 

Statement 1 examines whether the expression \(2^{513} + 2^{013} + 8^{13} + 3^{13}\) is divisible by 7. We will test each term independently using properties of exponents and modular arithmetic.

1. First, notice that \(8 = 2^3\), allowing us to rewrite the expression: \(2^{513} + 2^{013} + (2^3)^{13} + 3^{13}\\) simplifies to \(2^{513} + 2^{013} + 2^{39} + 3^{13}\).
2. According to Fermat's Little Theorem, for any integer \(a\) not divisible by a prime \(p\), \(a^{p-1} \equiv 1 \pmod{p}\).
3. Applying this theorem for each base modulo 7:
   • The cycle for \(2^n \mod 7\) is [2, 4, 1] (with a repeating pattern every 3 terms). Thus, \(2^1 \equiv 2 \mod 7\\),\(2^2 \equiv 4 \mod 7\\),\(2^3 \equiv 1 \mod 7\).
   • Finding each term,
     • \(2^{513} \equiv 2^{(3 \times 171) + 0} \equiv 1 \mod 7\)
     • \(2^{013} \equiv 2^{(3 \times 4) + 1} \equiv 2 \mod 7\)
     • \(2^{39} \equiv 1^{13} \equiv 1 \mod 7\)
   • Similarly, the cycle for \(3^n \mod 7\) is [3, 2, 6, 4, 5, 1] (with a repeating pattern every 6). Hence, \(3^{13} \equiv 3^1 \equiv 3 \mod 7\).

Summing these terms modulo 7, \((1+2+1+3) \equiv 7 \equiv 0 \mod 7.\) Thus, Statement 1 is correct.

Analysis of Statement 2:

Statement 2 assesses the parity (odd/even) of the integral part of \((7+4\sqrt{3})^{25}\).

1. Define \( a_n = (7 + 4\sqrt{3})^n + (7 - 4\sqrt{3})^n\).
2. The terms \((7 + 4\sqrt{3})\) and \((7 - 4\sqrt{3})\) are conjugates, causing cancels in irrational part of \(a_n\).
3. For large \(n\), the term \((7 - 4\sqrt{3})^n\) is negligible. Therefore, the dominant term \((7 + 4\sqrt{3})^n\) dictates the floor integer.
4. From the properties of \(a_n\), observe that \(a_n\) is always an integer satisfying \(a_n \equiv 2 \mod 4\), hence odd.

The validity of statement 2 being odd is confirmed. Therefore, Statement 2 is also correct.

Conclusion:

Both statements are correct. Thus, the correct answer is: Both Statements are correct.