Question:
Find the remainder when \(7^{103}\) is divided by 25.
\(7^{103}\) को 25 से विभाजित करने पर शेषफल ज्ञात कीजिए।
Options:
- 18
- 7
- 24
- 14
- 18
- 7
- 24
- 14
Correct Answer: a)
Explanation:
Using Euler’s Totient function \(\phi(25) = 25(1 – 1/5) = 20\).
Since \(\gcd(7, 25) = 1\), by Euler’s Totient Theorem:
\(7^{20} \equiv 1 \pmod{25}\).
\(103 = 20 \times 5 + 3\).
Therefore:
\(7^{103} = (7^{20})^5 \times 7^3 \equiv 1^5 \times 7^3 = 343 \pmod{25}\).
Now, \(343 = 25 \times 13 + 18\).
Hence, the remainder is 18.
Since \(\gcd(7, 25) = 1\), by Euler’s Totient Theorem:
\(7^{20} \equiv 1 \pmod{25}\).
\(103 = 20 \times 5 + 3\).
Therefore:
\(7^{103} = (7^{20})^5 \times 7^3 \equiv 1^5 \times 7^3 = 343 \pmod{25}\).
Now, \(343 = 25 \times 13 + 18\).
Hence, the remainder is 18.
यूलर प्रमेय द्वारा \(\phi(25) = 20\)
\(7^{20} \equiv 1 \pmod{25}\)
\(7^{103} = (7^{20})^5 \times 7^3 \equiv 1 \times 343 \pmod{25}\)
\(343 \div 25\) पर शेषफल 18 आता है।
\(7^{20} \equiv 1 \pmod{25}\)
\(7^{103} = (7^{20})^5 \times 7^3 \equiv 1 \times 343 \pmod{25}\)
\(343 \div 25\) पर शेषफल 18 आता है।