
問題の解答

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% 例題A4.2.14:$n$進法の位の数(One More)★★★
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自然数$N$を6進法と9進法で表すと,それぞれ3桁の数$abc_{(6)}$と$cab_{(9)}$になるとする.このとき$a,b,c$の値を求めよ.また,$N$を10進法で表せ.
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% 例題A4.2.14の解答(One More)★★★
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自然数$N$を6進法,9進法で表すと,それぞれ$abc_{(6)}$と$cab_{(9)}$であるから,
\[
1\leqq a\leqq 5,0\leqq b\leqq 5,1\leqq c\leqq 5
\]
$abc_{(6)},cab_{(9)}$をそれぞれ10進法で表すと,
\begin{align*}
&abc_{(6)}=a\cdot 6^2+b\cdot 6^1+c\cdot 6^0=36a+6b+c\cdots(\mathrm{i}),\\
&cab_{(9)}=c\cdot 9^2+a\cdot 9^1+b\cdot 9^0=81c+9a+b
\end{align*}
これらが等しいから,$36a+6b+c=81c+9a+b$
したがって,$27a=80c-5b$,すなわち,$27a=5(16c-b)\cdots(\mathrm{ii})$
これより,5と27は互いに素であるから,$a$は5の倍数である.
ゆえに,$1\leqq a\leqq 5$より,$a=5$
これを(ii)に代入して整理すると,$16c=b+27\cdots(\mathrm{iii})$
これより,$b+27$は16の倍数である.
したがって,$0\leqq b\leqq 5$より,$27\leqq b+27\leqq 32$であるから,$b+27=32$
ゆえに,$b=5$であり,(iii)より,$1\leqq c\leqq 5$を満たす整数$c$は,$c=2$
よって,$a=5,b=5,c=2$
また,この値を(i)に代入すると,$N=36\cdot 5+6\cdot 5+2=212$
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% 問題A4.2.14:$n$進法の位の数(One More)★★★
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自然数$N$を4進法と7進法で表すと,それぞれ2桁の数$ab_{(4)}$と$ba_{(7)}$になるとする.このとき$a,b$の値を求めよ.また,$N$を10進法で表せ.
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% 問題A4.2.14の解答(One More)★★★
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自然数$N$を4進法,7進法で表すと,それぞれ$ab_{(4)}$と$ba_{(7)}$であるから,
\[
1\leqq a\leqq 3,1\leqq b\leqq 3\cdots(\mathrm{i})
\]
$ab_{(4)},ba_{(7)}$をそれぞれ10進法で表すと,
\begin{align*}
&ab_{(4)}=a\cdot 4^1+b\cdot 4^0=4a+b\cdots(\mathrm{ii}),\\
&ba_{(7)}=b\cdot 7^1+a\cdot 7^0=7b+a
\end{align*}
これらが等しいから,$4a+b=7b+a$
したがって,$a=2b$
よって,(i)において,これを満たす$a,b$を求めると,$a=2,b=1$
また,この値を(ii)に代入すると,$N=4\cdot 2+1=9$
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