Exercise · ZW-FERMAT-S-LITTLE-THEOREM-009 · ★★☆ Intermediate
Compute \(6^{17}\bmod 17\).
Choose one option.
Show answer
- B. \(6\)
Exercise · ZW-FERMAT-S-LITTLE-THEOREM-009 · ★★☆ Intermediate
Choose one option.
Since \(17\) is prime, \(6^{17}\equiv 6^{1}\cdot 6^{16}\equiv 6\pmod{17}\), using \(a^p\equiv a\). The option \(1\) is a typical error from applying \(a^{p-1}\equiv 1\) to the exponent \(17\) instead of \(16\).
Fermat's little theorem states that if \(p\) is a prime number and \(a\) is an integer not divisible by \(p\), then \(a^{p-1}\equiv 1 \pmod{p}\). It is a basic tool for simplifying modular powers.
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