Its basically asking the remainder of \(\frac{63^{33}×36^{195}}{10}\)
And it all depends on the unit's digit of \(63^{33}×36^{195}\)
Units digit of \(63^{33}\) depends on cyclicity of 3.
\(63^{1}\) = 3
\(63^{2}\) = 9
\(63^{3}\) = 7
\(63^{4}\) = 1
\(63^{5}\) = 3
.
.
.
\(63^{32}\) = 1
\(63^{33}\) = 3
Hence the units digit would be:
3On the other hand, the units digit of \(36^{195}\), will depend on cyclicity of 6, which is 1 and the units digit it always 6.
Hence the unit's digit of the expression \(63^{33}×36^{195}\) is \(3x6=18\)
=>8
Therefore, the remainder when we divide the above expression by 10 would be 8
Option
D
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