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