The digit in the unit place of the number

(183!) + $3^{183}$ is

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Solution

The correct option is A
7

We know that n! terminates in 0 for n ≥ 5 and

$3^{4 n}$ terminator in 1, ( ∵ $3^{4}$ = 81 )

∴ $3^{180}$ = $(3^{4})^{45}$ terminates in 1

Also $3^{3}$ = 27 terminates in 7

∴ $3^{183}$ = $3^{180}$ $3^{3}$ terminates in 7.

∴ 183! + $3^{183}$ terminates in 7

i.e. the digit in the unit place = 7.

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