Mathematics
Senior High

log10底の3の2018乗について。(log10の3=0,4771)
3の2018乗の1の位の求め方について教えてください。

Answers

3^2018=3^(2×1009)
=9^1009
=9^(2^504+1)
=81^504×9=(80+1)^504×9
=(80k+1)×9 (kは自然数)
=720k+9

よって、3^2018の1のくらいの数は9

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