If \(a\) is a positive integer, prove that

\[ (2^{10})^a \bmod{100} = \begin{cases} 76, a \text{ even} \\ 24, a \text{ odd} \\ \end{cases} \]

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TopNewestThis techniques seem interesting. Could you illustrate this with an example?

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For example : 2^199 Solution: 16^119 =(2^4)^119 =2^796 =2^(10×79+6) =(2^10)^79+2^6 =24 × 64 (here a is odd,by applying the above rule) =4×4 =16 Hence 16 is the last digit place. Check it out by using calc.

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The last two digits of \(2^{199} \) is \(88\), and the last two digits of \(16^{119} \) is \(36\).

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Sorry question was mistyped by me the question is 16^199. Sry guys

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