問題の解答
検索用コード（LaTeX）
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% 例題A4.2.1：ユークリッドの互除法（One More）★
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(1) ユークリッドの互除法を用いて，357と544の最大公約数と最小公倍数を求めよ．
(2) ユークリッドの互除法を用いて，$\frac{638}{899}$を既約分数にせよ．
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% 例題A4.2.1の解答（One More）★
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(1)
\begin{align*}
544&=357\times 1+187\\
357&=187\times 1+170\\
187&=170\times 1+17\\
170&=17\times 10
\end{align*}
170と17の最大公約数は17であるから，357と544の最大公約数は，17
したがって，$357=21\times 17,544=32\times 17$
よって，357と544の最小公倍数は，
\[
21\times 32\times 17=11424
\]
(2)
\begin{align*}
899&=638\times 1+261\\
638&=261\times 2+116\\
261&=116\times 2+29\\
116&=29\times 4
\end{align*}
116と29の最大公約数は29であるから，638と899の最大公約数は，29
よって，$\frac{638}{899}=\frac{29\times 22}{29\times 31}=\frac{22}{31}$
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% 問題A4.2.1：ユークリッドの互除法（One More）★
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(1) ユークリッドの互除法を用いて，462と700の最大公約数と最小公倍数を求めよ．
(2) ユークリッドの互除法を用いて，$\frac{871}{1209}$を既約分数にせよ．
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% 問題A4.2.1の解答（One More）★
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(1)
\begin{align*}
700&=462\times 1+238\\
462&=238\times 1+224\\
238&=224\times 1+14\\
224&=14\times 16
\end{align*}
224と14の最大公約数は14であるから，462と700の最大公約数は，14
したがって，$462=33\times 14,700=50\times 14$
よって，462と700の最小公倍数は，
\[
33\times 50\times 14=23100
\]
(2)
\begin{align*}
1209&=871\times 1+338\\
871&=338\times 2+195\\
338&=195\times 1+143\\
195&=143\times 1+52\\
143&=52\times 2+39\\
52&=39\times 1+13\\
39&=13\times 3
\end{align*}
39と13の最大公約数は13であるから，871と1209の最大公約数は，13
よって，$\frac{871}{1209}=\frac{13\times 67}{13\times 93}=\frac{67}{93}$
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