現在,One Moreの数学II・B版を作成中です!

【数学A】例題4.1.6:最大公約数・最小公倍数2(One More)★★

【数学A】例題4.1.6:最大公約数・最小公倍数2(One More)
【数学A】例題4.1.6:最大公約数・最小公倍数2の例題ページ
問題の解答

【数学A】問題4.1.6:最大公約数・最小公倍数2の解答
検索用コード(LaTeX)
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A4.1.6:最大公約数・最小公倍数2(One More)★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

次の条件を満たす2つの自然数$a,b$の組をすべて求めよ.ただし,$a<b$とする.

(1) 和が150,最大公約数が10

(2) 積が180,最小公倍数が60

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 例題A4.1.6の解答(One More)★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) 最大公約数が10であるから,2つの自然数$a,b$は$a=10a^{\prime},b=10b^{\prime}$とおける.ただし,$a^{\prime},b^{\prime}$は互いに素な自然数であり,$a<b$より,$a^{\prime}<b^{\prime}$

和が150であるから,$10a^{\prime}+10b^{\prime}=150$

したがって,$a^{\prime}+b^{\prime}=15$

これを満たす,互いに素な自然数$a^{\prime},b^{\prime}$の組は,
\[
(a^{\prime},b^{\prime})=(1,14),(2,13),(4,11),(7,8)
\]

よって,

\[
(a,b)=(10,140),(20,130),(40,110),(70,80)
\]

(2) 最大公約数を$g$とすると,積が180,最小公倍数が60であるから,$180=g\cdot 60$

したがって,$g=3$であるから,$a=3a^{\prime},b=3b^{\prime}$とおける.ただし,$a^{\prime},b^{{\prime}}$は互いに素な自然数であり,$a<b$より,$a^{\prime}<b^{\prime}$

$60=3a^{\prime}b^{\prime}$が成り立つから,$a^{\prime}b^{\prime}=20$

これを満たす,互いに素な自然数$a^{\prime},b^{\prime}$の組は,
\[
(a^{\prime},b^{\prime})=(1,20),(4,5)
\]

よって,
\[
(a,b)=(3,60),(12,15)
\]

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A4.1.6:最大公約数・最小公倍数2(One More)★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

次の条件を満たす2つの自然数$a,b$の組をすべて求めよ.ただし,$a<b$とする.

(1) 和が180,最大公約数が15

(2) 積が400,最小公倍数が80

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% 問題A4.1.6の解答(One More)★★
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

(1) 最大公約数が15であるから,2つの自然数$a,b$は$a=15a^{\prime},b=15b^{\prime}$とおける.ただし,$a^{\prime},b^{\prime}$は互いに素な自然数であり,$a<b$より,$a^{\prime}<b^{\prime}$

和が180であるから,$15a^{\prime}+15b^{\prime}=180$

したがって,$a^{\prime}+b^{\prime}=12$

これを満たす,互いに素な自然数$a^{\prime},b^{\prime}$の組は,
\[
(a^{\prime},b^{\prime})=(1,11),(5,7)
\]

よって,$(a,b)=(15,165),(75,105)$

(2) 最大公約数を$g$とすると,積が400,最小公倍数が80であるから,$400=g\cdot 80$

したがって,$g=5$であるから,$a=5a^{\prime},b=5b^{\prime}$とおける.ただし,$a^{\prime},b^{{\prime}}$は互いに素な自然数であり,$a<b$より,$a^{\prime}<b^{\prime}$

$80=5a^{\prime}b^{\prime}$が成り立つから,$a^{\prime}b^{\prime}=16$

これを満たす,互いに素な自然数$a^{\prime},b^{\prime}$の組は,
\[
(a^{\prime},b^{\prime})=(1,16)
\]

よって,$(a,b)=(5,80)$

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