The last digit of $17^{256}$ is

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Solution

We have,
$17^{256} = (17^{2})^{128} = (289)^{128} = (290 - 1)^{128} =^{128} C_{0} (290)^{128} -^{128} C_{1} (290)^{127} + \dots -^{128} C_{127} (290) + 1 = 10 k + 1, k \in Z$

Hence, the last digit is $1$ .

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