方針の立て方

どれも10進法に直して処理し，その後，元の進数に直せばよい．

解答例

(72)(73)(74)……642
(75)(76)……14
(77)(78)……10
(79)(80)……12
(81)(82)……07
(83)(84)……15

解説

(1)
${2018}_{\left(10\right)}={642}_{\left(18\right)}$ ……(答)

(2)
${516}_{\left(n\right)}={5n^2+n+6}_{\left(10\right)}={2018}_{\left(10\right)}$
$\therefore5n^2+n+6=2018\Longleftrightarrow\left(n-14\right)\left(5n+71\right)=0\Leftrightarrow n=14,-\frac{71}{5}$
よって， $n=14$ ……(答)

(3)
$x^2-{22}_{\left(8\right)}x+{120}_{\left(8\right)}=0\Leftrightarrow x^2-{18}_{\left(10\right)}x+{80}_{\left(10\right)}=0\Leftrightarrow\left(x-8_{\left(10\right)}\right)\left(x-{10}_{\left(10\right)}\right)=0\Leftrightarrow x=8_{\left(10\right)},{10}_{\left(10\right)}\Leftrightarrow x={10}_{\left(8\right)},{12}_{\left(8\right)}$ ……(答)

(4)
${23}_{\left(m\right)}={2m+3}_{\left(10\right)}$
${114}_{\left(m\right)}={m^2+m+4}_{\left(10\right)}$
$\therefore x^2-{23}_{\left(m\right)}x+{114}_{\left(m\right)}=0\Longleftrightarrow x^2-\left(2m+3\right)_{\left(10\right)}x+\left(m^2+m+4\right)_{\left(10\right)}=0$
$5_{\left(m\right)}=5_{\left(10\right)}$ より，
${25}_{\left(10\right)}-\left(2m+3\right)_{\left(10\right)}\cdot5_{\left(10\right)}+\left(m^2+m+4\right)_{\left(10\right)}=0\Leftrightarrow m^2-9_{\left(10\right)}+{14}_{\left(10\right)}=0\Leftrightarrow\left(m-2_{\left(10\right)}\right)\left(m-7_{\left(10\right)}\right)=0\Leftrightarrow m=2_{\left(10\right)},7_{\left(10\right)}$
$5_{\left(m\right)}$ の表記が可能なのは， $m=7_{\left(10\right)}$ ……(答)
$\therefore x^2-\left(2\cdot7+3\right)_{\left(10\right)}x+\left(7^2+7+4\right)_{\left(10\right)}=0\Leftrightarrow x^2-{17}_{\left(10\right)}x+{60}_{\left(10\right)}=0\Leftrightarrow\left(x-5_{\left(10\right)}\right)\left(x-{12}_{\left(10\right)}\right)=0\Leftrightarrow x=5_{\left(10\right)},{12}_{\left(10\right)}=5_{\left(7\right)},{15}_{\left(7\right)}$
よって，求めるもう1つの解は， ${15}_{\left(7\right)}$ ……(答)