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What are the last 4 digits of 10 9 2017 109^{2017} ?


The answer is 6269.

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1 solution

Mark Hennings
Aug 2, 2018

Note that 10 9 2017 9 2017 + ( 2017 1 ) × 100 × 9 2016 9 2017 + 1700 × 9 2016 ( m o d 10000 ) 109^{2017} \; \equiv 9^{2017} + \binom{2017}{1}\times 100 \times 9^{2016} \; \equiv \; 9^{2017} + 1700 \times 9^{2016} \pmod{10000} Note that 9 1 ( m o d 4 ) 9 \equiv 1 \pmod{4} and 9 2 1 ( m o d 5 ) 9^2 \equiv 1 \pmod{5} , so that 9 10 1 ( m o d 25 ) 9^{10} \equiv 1 \pmod{25} , and hence 9 10 1 ( m o d 100 ) 9^{10} \equiv 1 \pmod{100} . Thus 10 9 2017 9 2017 + 1700 × 9 6 ( m o d 10000 ) 109^{2017} \equiv 9^{2017} + 1700 \times 9^6 \pmod{10000} Now 9 2 1 ( m o d 16 ) 9^2 \equiv 1 \pmod{16} , and 9 50 1 ( m o d 125 ) 9^{50} \equiv 1 \pmod{125} and 9 250 1 ( m o d 625 ) 9^{250} \equiv 1 \pmod{625} , and hence 9 250 1 ( m o d 10000 ) 9^{250} \equiv 1 \pmod{10000} . Thus we deduce that 10 9 2017 9 17 + 1700 × 9 6 ( m o d 10000 ) 109^{2017} \equiv 9^{17} + 1700 \times 9^6 \pmod{10000} and this last is small enough to evaluate by hand. We obtain that 10 9 2017 6269 ( m o d 10000 ) 109^{2017} \equiv \boxed{6269} \pmod{10000}

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