
問題の解答

検索用コード(LaTeX)
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% 例題A4.1.4:素因数の個数(One More)★★
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(1) $13!$が$2^k$で割り切れるとき,自然数$k$の最大値を求めよ.
(2) $60!$は,末尾に0が何個連続して並ぶか.
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% 例題A4.1.4の解答(One More)★★
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(1) 1から13までの自然数について,
2の倍数は6個,$2^2$の倍数は3個,$2^3$の倍数は1個
したがって,$13!$に含まれる因数2の個数は,
\[
6+3+1=10(\text{個})
\]
よって,求める自然数$k$の最大値は,$k=10$
(2) 求める0の個数は$60!$に含まれる因数10の個数に等しい.また,$10=2\cdot 5$であり,$60!$に含まれる因数5の個数が因数2の個数より少ないので,因数10の個数は因数5の個数に等しい.
1から60までの自然数について,
5の倍数は12個,$5^2$の倍数は2個
したがって,$60!$に含まれる因数5の個数は,
\[
12+2=14(\text{個})
\]
よって,求める0の個数は,14個
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% 問題A4.1.4:素因数の個数(One More)★★
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(1) $15!$が$2^k$で割り切れるとき,自然数$k$の最大値を求めよ.
(2) $50!$は,末尾に0が何個連続して並ぶか.
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% 問題A4.1.4の解答(One More)★★
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(1) 1から15までの自然数について,
2の倍数は7個,$2^2$の倍数は3個,$2^3$の倍数は1個
したがって,$15!$に含まれる因数2の個数は,$7+3+1=11(\text{個})$
よって,求める自然数$k$の最大値は,$k=11$
(2) 求める0の個数は$50!$に含まれる因数10の個数に等しい.また,$10=2\cdot 5$であり,$50!$に含まれる因数5の個数が因数2の個数より少ないので,因数10の個数は因数5の個数に等しい.
1から50までの自然数について,
5の倍数は10個,$5^2$の倍数は2個
したがって,$50!$に含まれる因数5の個数は,$10+2=12(\text{個})$
よって,求める0の個数は,12個
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検索用コード(LaTeX) 本文・解答側注 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % 基本事項A4.1.1:約数と倍数,素数と合成数(One More) %%%%%%%%%%%%%%...
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