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# help

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Find the remainder when 19^{90} is divided by 5.

May 11, 2020

#1
+21000
0

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.

May 12, 2020
#2
+24992
+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

May 12, 2020