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【数学A】例題4.1.5:最大公約数・最小公倍数1(One More)★

【数学A】例題4.1.5:最大公約数・最小公倍数1(One More)
【数学A】例題4.1.5:最大公約数・最小公倍数1の例題ページ
問題の解答

【数学A】問題4.1.5:最大公約数・最小公倍数1の解答
検索用コード(LaTeX)
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% 例題A4.1.5:最大公約数・最小公倍数1(One More)★
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(1) 次の各組の最大公約数と最小公倍数を求めよ.

(i) $198,276$

(ii) $450,630,840$

(2) $n$を正の整数とする.$n$と18の最小公倍数が72となるような$n$をすべて求めよ.

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% 例題A4.1.5の解答(One More)★
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(1)

(i) 与えられた2つの数を素因数分解すると,

\[
198=2\cdot 3^2\cdot 11,276=2^2\cdot 3\cdot 23
\]

最大公約数は,$2\cdot 3=6$

最小公倍数は,$2^2\cdot 3^2\cdot 11\cdot 23=9108$

(ii) 与えられた3つの数を素因数分解すると,

\[
450=2\cdot 3^2\cdot 5^2,630=2\cdot 3^2\cdot 5\cdot 7,840=2^3\cdot 3\cdot 5\cdot 7
\]

最大公約数は,$2\cdot 3\cdot 5=30$

最小公倍数は,$2^3\cdot 3^2\cdot 5^2\cdot 7=12600$

(2) 18,72をそれぞれ素因数分解すると,

\[
18=2\cdot 3^2,72=2^3\cdot 3^2
\]

したがって,18との最小公倍数が72である正の整数は,$2^3\cdot 3^a(a=0,1,2)$

ゆえに,求める正の整数$n$は,

\[
n=2^3\cdot 3^0,2^3\cdot 3^1,2^3\cdot 3^2
\]

よって,$n=8,24,72$

\[
\begin{array}{rrr}
2&)&198\\\cline{2-3}
3&)&99\\\cline{2-3}
3&)&33\\\cline{2-3}
&&11
\end{array}
\]

\[
\begin{array}{rrr}
2&)&276\\\cline{2-3}
2&)&138\\\cline{2-3}
3&)&69\\\cline{2-3}
&&23
\end{array}
\]

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% 問題A4.1.5:最大公約数・最小公倍数1(One More)★
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(1) 次の各組の最大公約数と最小公倍数を求めよ.

(i) $144,192$

(ii) $210,360,540$

(2) $n$を正の整数とする.$n$と20の最小公倍数が80となるような$n$をすべて求めよ.

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% 問題A4.1.5の解答(One More)★
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(1)

(i) 与えられた2つの数を素因数分解すると,

\[
144=2^4\cdot 3^2,192=2^6\cdot 3
\]

最大公約数は,$2^4\cdot 3=48$

最小公倍数は,$2^6\cdot 3^2=576$

(ii) 与えられた3つの数を素因数分解すると,

\[
210=2\cdot 3\cdot 5\cdot 7,360=2^3\cdot 3^2\cdot 5,540=2^2\cdot 3^3\cdot 5
\]

最大公約数は,$2\cdot 3\cdot 5=30$

最小公倍数は,$2^3\cdot 3^3\cdot 5\cdot 7=7560$

(2) 20,80をそれぞれ素因数分解すると,

\[
20=2^2\cdot 5,80=2^4\cdot 5
\]

したがって,20との最小公倍数が80である正の整数は,$2^4\cdot 5^a(a=0,1)$

ゆえに,求める正の整数$n$は,

\[
n=2^4\cdot 5^0,2^4\cdot 5^1
\]

よって,$n=16,80$

\[
\begin{array}{rrr}
2&)&144\\\cline{2-3}
2&)&72\\\cline{2-3}
2&)&36\\\cline{2-3}
2&)&18\\\cline{2-3}
3&)&9\\\cline{2-3}
&&3
\end{array}
\]

\[
\begin{array}{rrr}
2&)&192\\\cline{2-3}
2&)&96\\\cline{2-3}
2&)&48\\\cline{2-3}
2&)&24\\\cline{2-3}
2&)&12\\\cline{2-3}
2&)&6\\\cline{2-3}
&&3
\end{array}
\]

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