■階乗・二項係数の問題(その32)

[Q]1!+2!+3!+・・・+1000!  (mod10)

[A]  1+2+6+24+120+720+・・・

=1+2+6+24+Σ10k

=33+10n

1!+2!+3!+・・・+1000!=3  (mod10)

[Q]1!+2!+3!+・・・+1000!  (mod10^2)

[A]  1+2+6+24+120+720+・・・+9!+

=1+2+6+24+120+720+・・・+9!+Σ100k

=1+2+6+24+120+720+5040+40320+362880

 下3桁のみを計算すると・・・

=33+2080=2113

1!+2!+3!+・・・+1000!=13  (mod10^2)

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[Q](100!−99!−98!)/(100!+99!+98!)=?

[A]98!(100・99−99−1)/98!(100・99+99+1)

=(9900−99−1)/(9900+99+1)

=9800/10000=49/50

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