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

Statement 1:
Consider the expression: \[ 2^{513}+2^{013}+8^{13}+3^{13} \] Note that: \[ 8^{13} = (2^3)^{13} = 2^{39} \] Working modulo \(7\): \[ 2^3 \equiv 1 \pmod{7} \Rightarrow 2^{3k} \equiv 1 \pmod{7} \] \[ 2^{513} = 2^{3\cdot171} \equiv 1,\quad 2^{013} = 2^{13} = 2^{3\cdot4+1} \equiv 2,\quad 2^{39} = 2^{3\cdot13} \equiv 1 \] Also, \[ 3^6 \equiv 1 \pmod{7} \Rightarrow 3^{13} = 3^{6\cdot2+1} \equiv 3 \] Adding all: \[ 1+2+1+3 = 7 \equiv 0 \pmod{7} \] Hence, Statement 1 is correct
. Statement 2:
\[ (7+4\sqrt{3})^{25} \] Let: \[ a = 7+4\sqrt{3}, \quad b = 7-4\sqrt{3} \] Then: \[ ab = 49-48 = 1 \Rightarrow b=\frac{1}{a} \] Using binomial symmetry: \[ a^{25}+b^{25} \in \mathbb{Z} \] Also, \[ 0<b^{25}<1 \Rightarrow \lfloor a^{25} \rfloor = a^{25}+b^{25}-1 \] Since \(a^{25}+b^{25}\) is an integer and odd (sum of odd powers of conjugates), \[ \lfloor (7+4\sqrt{3})^{25} \rfloor \text{ is odd} \] Thus, Statement 2 is also correct
.
Final Conclusion:
Both statements are correct. \[ \boxed{\text{Option (1)}} \]