Find the remainder when 19^{90} is divided by 5.
Since we're dividing by 5, we need only to look at the last digit.
The last digit for each odd power of 19 is 9 -- when divided by 5, gives a remainder of 4.
The last digit for each even power of 19 is 1 -- when divided by 5, gives a remainder of 1.
Find the remainder when \(19^{90}\) is divided by \(5\).
\(\begin{array}{|rcll|} \hline && 19^{90} \pmod{5} \quad | \quad 19 \equiv 4 \equiv 4-5 \equiv -1 \pmod{5} \\ &\equiv& (-1)^{90} \pmod{5} \\ &\equiv& \mathbf{1} \pmod{5} \\ \hline \end{array}\)
The remainder is 1