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# number theory

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Find the last digit of $$\large 3^{14^{159}}$$

Jul 7, 2020

#1
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3^(14^159) mod 10^10 =.........9058119681 - The last digit is 1

Jul 7, 2020
#2
+8341
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Notice the pattern:

31 = 3

32 = 9

33 = 27

34 = 81

35 = 243

The pattern goes forever. The last digits will always be 3, 9, 7, 1, 3, 9, 7, 1, ...

That means if we have some ways to calculate which one of {3, 9, 7, 1} is the last digit, then we get the answer directly.

Because the pattern is periodic every 4 terms, we know that it is related to the power mod 4.

power mod 4 = 1 gives last digit = 3

power mod 4 = 2 gives last digit = 9

power mod 4 = 3 gives last digit = 7

power mod 4 = 0 gives last digit = 1

We attempt to find $$14^{159} \text{ mod } 4$$.

$$\quad 14^{159} \\\equiv 2^{159}\pmod 4\\\equiv 4^{79} \cdot 2 \pmod 4\\\equiv 0\pmod 4$$

That means the required last digit is 1.

Jul 7, 2020
#3
+25543
+2

Find the last digit of $$3^{\left(14^{159}\right)}$$

$$\begin{array}{|rcll|} \hline && \mathbf{3^{\left(14^{159}\right)} \pmod{10}} \\ &\equiv& 3^{\left(14^{158}*14\right)} \pmod{10} \\ &\equiv& 3^{\left(14^{158}*2*7\right)} \pmod{10} \\ &\equiv& 3^{\left(2*7*14^{158}\right)} \pmod{10} \\ &\equiv& \left(3^2\right)^{\left(7*14^{158}\right)} \pmod{10} \quad | \quad 3^2 \equiv -1 \pmod{10} \\ &\equiv& (-1)^{\left(7*14^{158}\right)} \pmod{10} \quad | \quad 7*14^{158}\ \text{is even}! \\ &\equiv& \mathbf{1} \pmod{10} \\ \hline \end{array}$$

The last digit of $$\large 3^{\left(14^{159}\right)}$$ is $$\mathbf{1}$$

Jul 8, 2020
edited by heureka  Jul 8, 2020