現在,One Moreの数学II・B版を作成中です!

【数学A】例題4.1.4:素因数の個数(One More)★★

【数学A】例題4.1.4:素因数の個数(One More)
【数学A】例題4.1.4:素因数の個数の例題ページ
問題の解答

【数学A】問題4.1.4:素因数の個数の解答
検索用コード(LaTeX)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A4.1.4:素因数の個数(One More)★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(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.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A4.1.4の解答(One More)★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(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

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A4.1.4:素因数の個数(One More)★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(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.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A4.1.4の解答(One More)★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(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

あわせて読みたい
【数学A】4章:数学と人間の活動(基本事項) 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) %%%%%%%%%%%%%%...
あわせて読みたい
【数学A】4章:数学と人間の活動(節末問題・章末問題) 節末A4.1.1〜A4.1.5の解答 節末A4.1.1節末A4.1.2節末A4.1.3節末A4.1.4節末A4.1.5 リンク(関連例題) https://onemath.net/onemorea-reidai4-1-3 https://onemath.net/o...
目次