Question

Given below are two statements: Statement I: \[ 25^{13}+20^{13}+31^{13} \text{ is divisible by } 7 \] Statement II: The integral part of \(\left(7+4\sqrt3\right)^{25}\) is an odd number. In the light of the above statements, choose the correct answer:

• A. Statement I is false but Statement II is true
• B. Statement I is true but Statement II is false
• C. Both Statement I and Statement II are false
• D. Both Statement I and Statement II are true

Hint:

Expressions of the form \((a+b\sqrt{n})^k\) are best handled using conjugates.

Answer 1

The Correct Option is D

Statement I: Work modulo \(7\): \[ 25\equiv4,\quad 20\equiv6,\quad 31\equiv3\pmod{7} \] Using Fermat’s theorem: \[ a^6\equiv1\pmod7 \Rightarrow a^{13}\equiv a\pmod7 \] Hence: \[ 25^{13}+20^{13}+31^{13}\equiv 4+6+3=13\equiv0\pmod7 \] \[ \Rightarrow \text{Statement I is true.} \] Statement II: \[ (7+4\sqrt3)(7-4\sqrt3)=49-48=1 \] Hence: \[ (7+4\sqrt3)^{25}+(7-4\sqrt3)^{25}\in\mathbb{Z} \] Since \(0<7-4\sqrt3<1\), \[ (7-4\sqrt3)^{25}\in(0,1) \] Thus, the integer part of \((7+4\sqrt3)^{25}\) equals: \[ (7+4\sqrt3)^{25}+(7-4\sqrt3)^{25}-1 \] which is clearly odd. \[ \Rightarrow \text{Statement II is true.} \] Final Conclusion: \[ \boxed{\text{Both Statement I and Statement II are true}} \]