Discussion

If \(2019^{19}\) is divided by 19, then what is the remainder ?

Answer: C

The remainder is 5.

First reduce the base modulo 19: 2019 = 19 × 106 + 5, so 2019 leaves remainder 5. Hence \(2019^{19}\) leaves the same remainder as \(5^{19}\).

Since 19 is prime and does not divide 5, Fermat's little theorem gives \(5^{18} \equiv 1 \pmod{19}\). Therefore:

\[ \begin{aligned} 5^{19} &= 5^{18} \times 5 \\ &\equiv 1 \times 5 = 5 \pmod{19} \end{aligned} \]

More generally \(a^{p} \equiv a \pmod{p}\) for a prime p, so the remainder is simply that of 2019 itself. The options 2, 3 and 7 do not satisfy this.

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