【数学A】4章：数学と人間の活動（節末問題・章末問題）

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% 節末問題A4.1.1：（One More）★★
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正の約数の個数が12個である自然数のうち，最も小さい数を求めよ．

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% 節末問題A4.1.1の解答（One More）★★
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正の約数の個数が12個である自然数は，次の4つの場合がある．

(i) 素数$p$を用いて$p^{11}$と表されるとき

このような自然数のうち，最も小さい数は，$2^{11}=2048$

(ii) 2つの異なる素数$p,q$を用いて$p^5q$と表されるとき

このような自然数のうち，最も小さい数は，$2^5\cdot 3=96$

(iii) 2つの異なる素数$p,q$を用いて$p^3q^2$と表されるとき

このような自然数のうち，最も小さい数は，$2^3\cdot 3^2=72$

(iv) 3つの異なる素数$p,q,r$を用いて$p^2qr$と表されるとき

このような自然数のうち，最も小さい数は，$2^2\cdot 3\cdot 5=60$

(i)〜(iv)より，求める自然数は，60

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% 節末問題A4.1.2：（One More）★★
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$n!$について，末尾に0がちょうど4個並ぶような最小の自然数$n$を求めよ．

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% 節末問題A4.1.2の解答（One More）★★
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$n!$の末尾に並ぶ0の個数は，$n!$に含まれる因数10の個数に等しい．

また，$10=2\cdot 5$であり，$n!$に含まれる素因数5の個数は素因数2の個数より少ないので，因数10の個数は素因数5の個数に等しい．

よって，$n!$に含まれる素因数5の個数がちょうど4となるような自然数$n$のうち，最小のものが求める数となるから，

\[
5\times 4=20
\]

より，求める自然数$n$は，20

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% 節末問題A4.1.3：（One More）★★
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分数$\frac{144}{35},\frac{234}{55}$のいずれに掛けても積が自然数となるような分数のうち，最小のものを求めよ．

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% 節末問題A4.1.3の解答（One More）★★
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与えられた条件を満たす分数を$\frac{a}{b}(a,b$は互いに素な自然数)とすると，$a$は35と55の公倍数，$b$は144と234の公約数である．

したがって，$\frac{a}{b}$が最小となるには，35と55の最小公倍数を$a$，144と234の最大公約数を$b$とすればよい．

よって，$35=5\cdot 7,55=5\cdot 11,144=2^4\cdot 3^2,234=2\cdot 3^2\cdot 13$であるから，

\[
a=5\cdot 7\cdot 11=385,b=2\cdot 3^2=18
\]

とすると，求める最小値は，$\frac{385}{18}$

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% 節末問題A4.1.4：（One More）★★★
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すべての自然数$n$について，$n$と$n+1$は互いに素であることを証明せよ．

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% 節末問題A4.1.4の解答（One More）★★★
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$n$と$n+1$が互いに素ではないと仮定すると，$n$と$n+1$は共通の素数$p$を約数にもち，

\[
n=ap,n+1=bp
\]

とおける．ただし，$a,b$は自然数である．

このとき，

\[
(n+1)-n=bp-ap
\]

すなわち，$1=(b-a)p$

ここで，$b-a$は整数であるから，$p$は1の約数である．

これは$p$が素数であることに矛盾する．

よって，$n$と$n+1$は互いに素である．$\blacksquare$

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% 節末問題A4.1.5：（One More）★★★
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$\frac{n}{196}$が1より小さい既約分数となるような正の整数$n$は全部で何個あるか．

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% 節末問題A4.1.5の解答（One More）★★★
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$n$は196と互いに素である195以下の正の整数である．$196=2^2\cdot 7^2$であるから，このような数は，1から195までの195個の自然数のうち，2または7の倍数を除いたものである．1から195までの自然数について，

2の倍数の個数は，$1\cdot 2,2\cdot 2,\ldots,97\cdot 2$の$97$個

7の倍数の個数は，$1\cdot 7,2\cdot 7,\ldots,27\cdot 7$の$27$個

14の倍数の個数は，$1\cdot 14,2\cdot 14,\ldots,13\cdot 14$の$13$個

よって，求める個数は，

\[
195-(97+27-13)=84(\text{個})
\]

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% 節末問題A4.2.1：（One More）★★
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120と168の最大公約数$g$を求め，$g=120m+168n$となる整数$m,n$の組を1つ求めよ．

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% 節末問題A4.2.1の解答（One More）★★
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120と168について，ユークリッドの互除法を用いる．

\[
\begin{array}{ll}
168=120\times 1+48\text{より，}&168-120\times 1=48\cdots(\mathrm{i})\\
120=48\times 2+24\text{より，}&120-48\times 2=24\cdots(\mathrm{ii})\\
48=24\times 2&
\end{array}
\]

よって，最大公約数$g$は，$g=24$

(ii)に(i)を代入すると，$120-(168-120\times 1)\times 2=24$より，$120\times 3-168\times 2=24$

よって，求める整数$m,n$の組の1つは，$(m,n)=(3,-2)$

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% 節末問題A4.2.2：（One More）★★★
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方程式$19x+53y=7$を満たす整数の組$(x,y)$の中で，$|x-y|$が最小となるものを求めよ．

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% 節末問題A4.2.2の解答（One More）★★★
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方程式$19x+53y=7\cdots(\mathrm{i})$の係数である$19$と$53$について，ユークリッドの互除法を用いる．

\[
\begin{array}{ll}
53=19\times 2+15\text{より，}&53-19\times 2=15\cdots(\mathrm{ii})\\
19=15\times 1+4\text{より，}&19-15\times 1=4\cdots(\mathrm{iii})\\
15=4\times 3+3\text{より，}&15-4\times 3=3\cdots(\mathrm{iv})\\
4=3\times 1+1\text{より，}&4-3\times 1=1\cdots(\mathrm{v})
\end{array}
\]

(v) に(iv)を代入すると，$4-(15-4\times 3)\times 1=1$より，

\[
4\times 4+15\times(-1)=1
\]

これに(iii)を代入すると，$(19-15\times 1)\times 4+15\times(-1)=1$より，

\[
19\times 4+15\times(-5)=1
\]

これに(ii)を代入すると，$19\times 4+(53-19\times 2)\times(-5)=1$より，

\[
53\times(-5)+19\times 14=1
\]

この両辺に$7$を掛け合わせると，$53\times(-35)+19\times 98=7\cdots(\mathrm{vi})$

$(\mathrm{i})-(\mathrm{vi})$より，$19(x-98)+53(y+35)=0$

したがって，$19(x-98)=-53(y+35)\cdots(\mathrm{vii})$

19と53は互いに素であるから，$x-98$は53の倍数となり，$k$を整数とすると，$x-98=53k$，すなわち，$x=53k+98$

これを(vii)に代入して整理すると，$y=-19k-35$

ゆえに，(i)を満たす整数の組$(x,y)$は，$\{\begin{array}{l}x=53k+98\\
y=-19k-35\end{array}(k\text{は整数})\cdots(\mathrm{viii})$

したがって，$|x-y|=|72k+133|$

これが最小となる$k$の値は，$k=-2$

よって，求める整数の組$(x,y)$は，(viii)より，$(x,y)=(-8,3)$

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% 節末問題A4.2.3：（One More）★★
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ある自然数から35を引いた数と，36を加えた数がともに平方数となった．このとき，その自然数を求めよ．

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% 節末問題A4.2.3の解答（One More）★★
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求める自然数を$n$とする．

$n$から35を引いた数，$n$に36を足した数はともに平方数となるから，

\[
n-35=p^2\cdots(\mathrm{i}),n+36=q^2\cdots(\mathrm{ii})
\]

とおける．ただし，$p,q$は自然数とする．

ここで，$n-35<n+36$より，$p^2<q^2$，すなわち，$p<q$

(ii)$-$(i)より，$q^2-p^2=71$

したがって，$(q+p)(q-p)=71$

$q+p>q-p>0$より，

\[
\{\begin{array}{l}q+p=71\\
q-p=1\end{array}
\]

これを解くと，$p=35,q=36$

ゆえに，(i)に代入すると，$n-35=35^2$

したがって，$n=1260$

よって，求める自然数は，$1260$

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% 節末問題A4.2.4：（One More）★★
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$n$を5以上の整数とする．

(1) 十進法で表された数$(n+1)^2$を$n$進法で表せ．

(2) 十進法で表された数$(2n-1)^2$を$n$進法で表したとき，$n$の位の数を求めよ．

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% 節末問題A4.2.4の解答（One More）★★
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(1) $(n+1)^2=n^2+2n+1=1\times n^2+2\times n+1=121_{(n)}$

(2)

\begin{align*}
(2n-1)^2&=4n^2-4n+1\\
&=3n^2+n^2-4n+1\\
&=3n^2+(n-4)n+1\\
&=3\times n^2+(n-4)\times n+1
\end{align*}

よって，$n$進法で表したときの$n$の位の数は，$n-4$

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% 節末問題A4.2.5：（One More）★★
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十進法の1440を$n$進法で表すと$10400_{(n)}$になった．$n$の値を求めよ．

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% 節末問題A4.2.5の解答（One More）★★
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$10400_{(n)}$を十進法で表すと，

\[
10400_{(n)}=1\times n^4+0\times n^3+4\times n^2+0\times n+0=n^4+4n^2
\]

これが1440と等しくなるので，$n^4+4n^2=1440$

したがって，$n^4+4n^2-1440=0$

ゆえに，

\[
(n^2+40)(n^2-36)=0
\]

よって，$n$は5以上の自然数であるから，$n=6$

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% 章末問題A4.1：（One More）★★★
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${}_{80}\mathrm{C}_{40}$が$2^n$で割り切れるとき，自然数$n$の最大値を求めよ．

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% 章末問題A4.1の解答（One More）★★★
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与えられた${}_{80}\mathrm{C}_{40}$は，${}_{80}\mathrm{C}_{40}=\frac{80!}{40!40!}$と表される．

1から40までの自然数について，

2の倍数は20個，4の倍数は10個，8の倍数は5個，

16の倍数は2個，32の倍数は1個

したがって，$40!$に含まれる因数2の個数は，

\[
20+10+5+2+1=38(\text{個})
\]

また，1から80までの自然数について，

2の倍数は40個，4の倍数は20個，8の倍数は10個，

16の倍数は5個，32の倍数は2個，64の倍数は1個

ゆえに，$80!$に含まれる因数2の個数は，

\[
40+20+10+5+2+1=78(\text{個})
\]

したがって，${}_{80}\mathrm{C}_{40}$に含まれる因数2の個数は，

\[
78-38\times 2=2(\text{個})
\]

よって，求める自然数$n$の最大値は，$n=2$

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% 章末問題A4.2：（One More）★★
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$n$を自然数とする．$n+4$は5の倍数であり，$n+9$は11の倍数である．このような自然数$n$で300より小さいものは何個あるか．

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% 章末問題A4.2の解答（One More）★★
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$n+4,n+9$は，それぞれ$n+4=5k\cdots(\mathrm{i}),n+9=11l\cdots(\mathrm{ii})(k,l\text{は自然数})$とおける．

(ii)$-$(i)より，$5=11l-5k$

したがって，$11l=5(k+1)$

11と5は互いに素であるから，$l$は5の倍数である．

ゆえに，$l=5m(m\text{は自然数})$と表せる．

(ii)より，$n+9=11l=11\cdot 5m=55m$であるから，$n=55m-9$

したがって，$1\leqq n<300$より，$1\leqq 55m-9<300$であるから，

\[
\frac{10}{55}\leqq m<\frac{309}{55}
\]

$m$は自然数であるから，$1\leqq m\leqq 5$

よって，これを満たす自然数$m$は5個であるから，条件を満たす自然数$n$は，5個

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% 章末問題A4.3：（One More）★★★★
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(1) 2つの自然数$a$と$b(a>b)$が互いに素であるとき，$a$と$a-b$も互いに素であることを証明せよ．

(2) 504以下の自然数で，504と互いに素な自然数の個数を求めよ．

(3) 504以下の自然数で，504と互いに素な自然数の総和を求めよ．

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% 章末問題A4.3の解答（One More）★★★★
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(1) $a$と$a-b$が互いに素ではないと仮定すると，$a$と$a-b$はある素数$p$を約数にもち，$a=pm\cdots\text{(i)},a-b=pn\cdots\text{(ii)}$とおける．ただし，$m,n$は自然数とする．

$(\mathrm{i})-(\mathrm{ii})$より，$b=p(m-n)$

ここで，$m-n$は整数であるから，$p$は$b$の約数でもある．

したがって，$p$は$a$と$b$の公約数となり，これは$a$と$b$が互いに素であることに矛盾する．

よって，$a$と$a-b$は互いに素である．$\blacksquare$

(2) $504=2^3\times 3^2\times 7$であるから，504は素因数として$2,3,7$をもつ．

504以下の自然数について，

2の倍数は$252$個，3の倍数は$168$個，7の倍数は$72$個

6の倍数は$84$個，21の倍数は$24$個，14の倍数は$36$個，42の倍数は$12$個

したがって，504以下の自然数で，504と互いに素ではない自然数の個数は，

\[
252+168+72-84-24-36+12=360(\text{個})
\]

よって，504以下の自然数で，504と互いに素な自然数の個数は，

\[
504-360=144(\text{個})
\]

(3) 504以下の自然数で504と互いに素な自然数は，

\[
1,5,11,13,\ldots,491,493,499,503
\]

の計144個である．

ここで，(1)より，504以下の自然数の1つを$n$とすると，504と$n$が互いに素であるとき，$504-n$と504も互いに素である．

したがって，504以下の自然数で，504と互いに素な自然数は，和が504となる2つの数の組に分けることができ，その組の数は，

\[
144\div 2=72(\text{組})
\]

よって，504以下の自然数で，504と互いに素な自然数の総和は，

\[
504\times 72=36288
\]

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% 章末問題A4.4：（One More）★★★
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$x$についての2次方程式$x^2+2ax+2a-8=0$が異なる2つの整数解をもつような整数$a$の値を求めよ．

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% 章末問題A4.4の解答（One More）★★★
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与えられた2次方程式の判別式を$D$とすると，

\[
\frac{D}{4}=a^2-(2a-8)=a^2-2a+8=(a-1)^2+7
\]

したがって，この2次方程式は，$a$の値に関わらず，異なる$2$つの実数解をもつ．

この2次方程式の解は，
\[
x=-a\pm\sqrt{(a-1)^2+7}
\]

方程式が異なる2つの整数解をもつとき，$\sqrt{(a-1)^2+7}$は整数となるから，$\sqrt{(a-1)^2+7}=b$（$b$は正の整数）とおける．

両辺を2乗して整理すると，
\[
(a-1)^2-b^2=-7
\]

ゆえに，
\[
\{(a-1)+b\}\{(a-1)-b\}=-7
\]

ここで，$b>0$より，$a-1+b>a-1-b$であるから，

\[
(a-1+b,a-1-b)=(7,-1),(1,-7)
\]

したがって，

\[
(a,b)=(4,4),(-2,4)
\]

よって，
\[
a=4,-2
\]

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% 章末問題A4.5：（One More）★★★★
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6の約数$1,2,3,6$の和は6の2倍になっている．このように，正の約数の和がその数の2倍に等しいとき，その数を完全数という．$p,q$を異なる素数として，次の問いに答えよ．

(1) $pq$の形の完全数をすべて求めよ．

(2) $p^2q$の形の完全数をすべて求めよ．

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% 章末問題A4.5の解答（One More）★★★★
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(1) $pq$の約数は，$1,p,q,pq$の4個であるから，$pq$が完全数であるための条件は，

\[
1+p+q+pq=2pq
\]

したがって，$(p-1)(q-1)=2$

ここで，$p\geqq 2,q\geqq 2$であるから，

\[
(p-1,q-1)=(1,2),(2,1)
\]

ゆえに，$(p,q)=(2,3),(3,2)$

よって，$pq$の形の完全数は，$6$

(2) $p^2q$の約数は，$1,p,q,p^2,pq,p^2q$の6個であるから，$p^2q$が完全数であるための条件は，

\[
1+p+q+p^2+pq+p^2q=2p^2q
\]

したがって，$(1+p+p^2)(1+q)=2p^2q\cdots(\mathrm{i})$

ここで，$1+p+p^2$は$p$で割り切れないから，$1+q$は$p$で割り切れる．

ゆえに，(i)を，

\[
(1+p+p^2)\frac{1+q}{p}=2pq
\]

と式変形すると，$1+p+p^2$は同様に$p$で割り切れないから，$\frac{1+q}{p}$は$p$で割り切れる．

また，(i)の両辺を$p^2$で割ると，

\[
(1+p+p^2)\frac{1+q}{p^2}=2q
\]

$1+p+p^2$は奇数であるから，$\frac{1+q}{p^2}$は2で割り切れる．

ゆえに，

\[
1+q=2p^2\cdot r(r\text{は整数})\cdots(\mathrm{ii})
\]

とおける．これを(i)に代入すると，

\[
(1+p+p^2)r=q\cdots(\mathrm{iii})
\]

ここで，$q$は素数であるから，$r=1$であり，(ii)，(iii)に代入すると，

\[
1+q=2p^2,1+p+p^2=q
\]

これを解くと，$p=2,q=7$

よって，$p^2q$の形の完全数は，$28$

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% 節末問題A4.1.1の側注：（One More）★★
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% 側注1
$p^{11}$の正の約数の個数は，

\[
11+1=12\text{(個)}
\]

% 側注2
$p^5q$の正の約数の個数は，

\[
(5+1)(1+1)=12\text{(個)}
\]

% 側注3
$p^3q^2$の正の約数の個数は，

\[
(3+1)(2+1)=12\text{(個)}
\]

% 側注4
$p^2qr$の正の約数の個数は，

\[
(2+1)(1+1)(1+1)=12\text{(個)}
\]

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% 節末問題A4.1.2の側注：（One More）★★
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% 側注1
末尾に並ぶ0の個数は，因数10の個数と一致する．

% 側注2
$n!$は素因数5をちょうど4個含む数となる．

% 側注3
$20<5^2$であるから，$5^2$の倍数を考える必要はない．

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% 節末問題A4.1.3の側注：（One More）★★
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% 側注1
$\frac{144}{35}\times\frac{a}{b}$が自然数となるとき，$a$は35の倍数であり，$b$は144の約数である．また，$\frac{234}{55}\times\frac{a}{b}$が自然数となるとき，$a$は55の倍数であり，$b$は234の約数である．

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% 節末問題A4.1.4の側注：（One More）★★★
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% 側注1
背理法を用いる．$n$と$n+1$が互いに素ではない$\Leftrightarrow n$と$n+1$が素数を公約数にもつ

% 側注2
1の約数は，$1,-1$であり，どちらも素数ではない．

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% 節末問題A4.1.5の側注：（One More）★★★
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% 側注1
$195=2\times 97+1$

% 側注2
$195=7\times 27+6$

% 側注3
$195=13\times 14+13$

% 側注4
オイラー関数を用いると，$\phi(196)=\phi(2^2\cdot 7^2)=196(1-\frac{1}{2})(1-\frac{1}{7})=84$のように検算できる．

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% 節末問題A4.2.1の側注：（One More）★★
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% 側注1
$48=168-120\times 1$

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% 節末問題A4.2.2の側注：（One More）★★★
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% 側注1
$3=15-4\times 3$

% 側注2
$4=19-15\times 1$

% 側注3
$15=53-19\times 2$

% 側注4
$x=98,y=-35$は方程式$19x+53y=7$を満たす整数解の1つである．

% 側注5
$k=-2$のとき，$|x-y|=11$

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% 節末問題A4.2.3の側注：（One More）★★
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% 側注1
71は素数である．

% 側注2
$0<p<q$より，
\[
q+p>q-p
\]

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% 節末問題A4.2.4の側注：（One More）★★
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% 側注1
$n$の位の数が0以上$n-1$以下の整数となるように，$(n-4)n$と式変形する．

% 側注2
$n$は5以上の整数であるから，$n-4$は$0\leqq n-4\leqq n-1$を満たす．

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% 節末問題A4.2.5の側注：（One More）★★
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% 側注1
$n^2=t$とおくと，
\[
t^2+4t-1440=0
\]

これを因数分解すると，
\[
(t+40)(t-36)=0
\]

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% 章末問題A4.1の側注：（One More）★★★
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% 側注1
$80!$，$40!$に含まれる因数2の個数をそれぞれ$p,q$とすると，2を因数にもたない自然数$u,v$を用いて，

\[
80!=u\cdot 2^p,40!=v\cdot 2^q
\]

と表される．このとき，

\[
\frac{80!}{40!\cdot 40!}=\frac{u}{v^2}\cdot 2^{p-2q}
\]

よって，${}_{80}\mathrm{C}_{40}$に含まれる因数2の個数は，$p-2q$個となる．

% 側注2
${}_{80}\mathrm{C}_{40}$に含まれる因数2の個数は，$(p-2q)$個

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% 章末問題A4.2の側注：（One More）★★
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% 側注1
$a,b$は互いに素で，$ak$が$b$の倍数であるならば，$k$は$b$の倍数である$(a,b,k$は整数)．

% 側注2
$\frac{309}{55}=5.618\ldots$

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% 章末問題A4.3の側注：（One More）★★★★
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% 側注1
背理法を用いる．

% 側注2
例えば，$504\div 2=252(\text{個})$

% 側注3
最小公倍数の倍数を考える．

% 側注4
2または3または7の倍数である．

% 側注5
\begin{align*}
&n(A\cup B\cup C)\\
=&n(A)+n(B)+n(C)\\
-&n(A\cap B)-n(B\cap C)\\
-&n(C\cap A)+n(A\cap B\cap C)
\end{align*}

% 側注6
$1+503=504,5+499=504,11+493=504,\ldots$と和が504になっていることに注目する．

% 側注7
$(1,503),(5,499),\ldots$のように，和が504となる2つの数の組が72組できる．

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% 章末問題A4.4の側注：（One More）★★★
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% 側注1
$x$の係数が偶数であるから，
\[
x=\frac{-b^{\prime}\pm\sqrt{b^{\prime 2}-ac}}{a}
\]
を用いるとよい．

% 側注2
$a$は整数，$b$は正の整数である．また，$\sqrt{(a-1)^2+7}$が整数であれば，$x=-a\pm\sqrt{(a-1)^2+7}$は整数となるから，逆を確かめなくてよい．

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% 章末問題A4.5の側注：（One More）★★★★
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% 側注1
移項して，因数分解する．

% 側注2
$q+pq+p^2q=q(1+p+p^2)$より，$(1+p+p^2)(1+q)$と因数分解できる．

% 側注3
$1+p+p^2$を$p$で割ると1余るので，$p$で割り切れない．

% 側注4
$1+p+p^2=1+p(p+1)$は奇数であるから，2で割り切れない．

% 側注5
両辺を$2p^2\neq 0$で割り，整理している．

% 側注6
$p\geqq 2,q\geqq 2$である．