Question 110

If $$3^{48} = x(mod 10)$$ and if $$0 \leq x \leq 9$$ then $$x =$$

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$$3^{48} = x(mod 10)$$

We need to find remainder when $$3^{48\ }$$ is divided by 10.

3 has a cyclicity if 4. So as 48 is a multiple of 4, so last digit will be 1.

Hence remainder on division  by 10 =1

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