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# Unit digit and power

If $$a$$ is a positive integer, prove that

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

Note by Mafia MaNiAc
9 months, 3 weeks ago

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This techniques seem interesting. Could you illustrate this with an example? · 9 months, 3 weeks ago

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. · 9 months, 2 weeks ago

Sorry question was mistyped by me the question is 16^199. Sry guys · 9 months, 2 weeks ago

The last two digits of $$2^{199}$$ is $$88$$, and the last two digits of $$16^{119}$$ is $$36$$. · 9 months, 2 weeks ago

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