The remainder when the number 3 2 2 3 - (3 2 2)3 is divide by 8, is

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Solution

Q.
The remainder when the number $3^{2^{2^{3}}} - {(3^{2^{2}})}^{3}$ is divide by 8, is ______

Ans.
(0)

$3^{256} - 3^{12} = 3^{12} \times (3^{256 - 12} - 1) = {(1 + 8)}^{6} ({(1 + 8)}^{112} - 1) = (1 + 8 λ) (1 + 8 μ - 1)$

$= 8 μ (1 + 8 λ), which is divisible by 8 . Hence, remainder is zero.$

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