If $α, β, γ$ are the last three digits of $17^{256}$ then the value of $α + β + γ$ is

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answer is C.

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Detailed Solution

$\begin{array}{l} 17^{256} = {(17^{2})}^{128} = (289)^{128} = (290 - 1)^{128} \\ ∴ 17^{256} = .^{128} C_{0} (290)^{128} - .^{128} C_{1} (290)^{127} + .^{128} C_{2} (290)^{126} - \dots . . \\ - . .^{128} C_{125} (290)^{3} + .^{128} C_{126} (290)^{2} - . .^{128} C_{127} (290) + 1 \\ = [.^{128} C_{0} (290)^{128} - .^{128} C_{1} (290)^{127} + .^{128} C_{2} (290)^{126} - \dots . . \\ - .^{128} C_{125} (290)^{3}] + .^{128} C_{126} (290)^{2} - . .^{128} C_{127} (290) + 1 \end{array}$

$\begin{array}{l} = 1000 m + \frac{(128) (127)}{2} (290)^{2} - 128 \times 290 + 1 \\ = 1000 m + (128) (127) (290) (145) - 128 \times 290 - 1 \\ = 1000 m + (128) (290) (127 \times 145 - 1) + 1 \\ = 1000 m + (128) (290) (18414) + 1 \\ = 1000 m + 683527680 + 1 \\ = 1000 m + 683527000 + 680 + 1 \\ = 1000 (m + 683527) + 681 \end{array}$