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== 約数の個数,約数の総和(入試問題) ==

【約数の個数】
 正の整数Nの(正の)約数の個数は,Nを素因数分解することにより求めることができる.
 すなわち,p, q, rを素数とし,a, b, cを正の整数とするとき,Nの素因数分解が
N=paqbrc
になるとき,Nの(正の)約数の個数は,
(a+1)(b+1)(c+1)個になる.
(解説)
例えば,12=2231の約数は,112自身も含めて,
1=2030, 2=2130, 4=2230,
3=2031, 6=2131, 12=2231
の6個になる.これは,12=2231の約数が
2m3n
の形で表され,m=0, 1, 2およびn=0, 1という形で各指数が0の場合を1通りとして数えるからである.
(2+1)(1+1)=6個になる.
※約数の個数は,素因数分解したときの,素数p, q, rではなく,その指数a, b, cによって決まることに注意
【約数の総和】
 正の整数Nの素因数分解が
N=paqbrc
になるとき,Nの(正の)約数の総和は,
(1+p+・・・+pa)(1+q+・・・+qb)(1+r+・・・+rc)
になる.
 この公式は,数学Bの等比数列の和の公式を使えば,次の形に書ける.
pa+11p1×qb+11q1×rc+11r1
(解説)
例えば,12=2231に対して
(20+21+22)(30+31+31)
という式を展開したものを考えると
2030+2130+2230
+2031+2131+2231
となって「約数の総和」が得られる.(自分で思いつくのは,難しいかもしれないが,言われれば分かる話である)
 一般に,
(1+p+・・・+pa)(1+q+・・・+qb)(1+r+・・・+rc)
を展開すると,
各々のかっこから1つの項を選んで掛けたのものの和になる.(この式のかっこを外すと,約数が1回ずつ出てくる)
 そこで,この式が約数の総和に等しい.
【例1】 18=2132の約数の総和は
(1+2)(1+3+9)=39
【例2】 72=2332の約数の総和は
(1+2+4+8)(1+3+9)=195
【例3】 8=23の約数の総和は
1+2+4+8=15
※大学入試では,この公式に数字を入れるだけで答が出るような問題は,めったに出ない.

【問題1】
 2016の正の約数は全部で個あり,それらの平均はである.
(2016年度慶応義塾大理工学部)
[解答を見る]

【問題2】
 2016の正の約数の中で,偶数であるものの個数は 1 個で,これらすべての和は 2 である.
(2016年度福岡大医.理.工学部)
[解答を見る]

【問題3】
 12nの正の約数の個数が28個となるような自然数nは,n=である.
(2014年度慶応義塾大看護医療学部)
[解答を見る]

【問題4】
(1) 約数の個数が18である最小の自然数mを求めよ.
(2) 自然数nの約数の個数が60で,かつ1, 2, 3, 4, 5, 6, 7, 8, 9, 10がnの約数であるとき,nを求めよ.
(2000年度産業医大)
[解答を見る]

【問題5】
(1) 108の正の約数の個数を求めよ.
(2) a, b, c, dを自然数とし,a≧cとする.m=2a3b, n=2c3dについて,m, nの正の約数の個数がそれぞれ80, 72で,mnの正の公約数の個数が45であるという.このとき,a, b, c, dを求めよ.
(2005年度群馬大教育学部)
[解答を見る]

【問題6】
 nを正の整数とする.10nの正の約数すべての積は
である.
(2011年度早稲田大商学部)
[解答を見る]

(参考)
 自然数nの(正の)約数の総和をS(n)で表すとき,
S(n)=2nとなるときnを完全数という.
S(n)>2nとなるときnを過剰数という.
S(n)<2nとなるときnを不足数という.
【例】 n=6, 28, 496, 8128, ... は完全数である.
6=2×3=21(221)・・・(*1)
約数の和は(1+2)+(3+6)=12
28=4×7=22(231)
約数の和は(1+2+4)+(7+14+28)=56
496=16×31=24(251)
約数の和は
(1+2+4+8+16)+(31+62+124+248+496)=992
8128=64×127=26(271)
約数の和は
(1+2+4+8+16+32+64)+(127+254+508+1013+2032+4064+8128)=16256
 一般に2m1が素数であるとき,2m1(2m1)の形に書ける自然数は完全数である.(ユークリッド言論)・・・(*2)
 オイラーは,偶数の完全数は上記の形のものだけであることを証明した(1772)
2m1が素数であるためには,mが素数でなければならない.
mが素数であって,Mm=2m1も素数となるとき,Mmはメルセンス素数と呼ばれる.
• 奇数の完全数は見つかっていない.あるかないかも証明されていない.

==高校数学で♪~楽しく~♪解けそうな問題==
 「完全数,過剰数,不足数」の定義を簡単に示しておけば,次の各問題は大学入試に出題可能なレベルです.
【追加問題1】
 p, qを異なる素数とするとき,N=pqの形に書ける完全数を求めてください.
(筆者作成・・・間違いがあればお知らせください)
(解答)
 N=pqにおいて,p, qは素数であるから,約数の総和は
(1+p)(1+q)
これが完全数であるには
(1+p)(1+q)=2pq
1+p+q+pq=2pq
(p−1)(q−1)=2
p, qの対称式だから,p<qと仮定してもよい
このとき
p−1=1, q−1=2
p=2, q=3
N=pq=6
【追加問題2】
 2m1が素数であるとき,2m1(2m1)の形に書ける自然数は完全数であることを証明してください.
(筆者作成・・・間違いがあればお知らせください)
(解答)
 p=2m1が素数であるから,N=2m1(2m1)の形に書ける自然数の約数は
1,2,22,23,,2m1,p
約数の総和は
2m121×(p+1)=(2m1)2m=2N
よって,Nは完全数

【追加問題3】・・・上記【問題5】の類似問題
 偶数の完全数n=2p1(2p1)の約数の積がnpで与えられることを示せ.
「初等整数論9章」Lames J.Tattersall原著
小松尚夫訳 / 森北出版 P.132
(解答)
 n=2p1(2p1)が完全数であるときは,q=2p1は素数である.
 このとき,nの約数は
1,2,22,23,,2p1
q,2q,22q,23q,,2p1q
であるから,それらの積は
1×2×22×23××2p1
×q×2q×22q×23q××2p1q
=21+2+3++(p1)
×qp×21+2+3++(p1)
=2(p1)p2
×qp×2(p1)p2
=2(p1)pqp
=(2(p1)q)p
=np

【追加問題4】
 奇数の不足数や偶数の不足数が無数に存在することを示せ.
「初等整数論9章」Lames J.Tattersall原著
小松尚夫訳 / 森北出版 P.131
(解答)
 奇数の素数が無限に存在することは既知とする.
 奇数の素数p (>2)の約数は1, pであるから,それらの約数の総和は,1+pになり,
2p−S(p)=2p−(p+1)=p−1>0
だから,pは不足数である.したがって,奇数の不足数は無数に存在する.
 奇数の素数p (>5)に対して,n=2pという偶数を考えると,その約数は,1, 2, p, 2pであるから,それらの約数の総和は,S(n)=1+2+p+2p=3p+3=3(p+1)になり,
2n−S(n)=4p−3(p+1)
=p−3>0
であるから,nは不足数である.
 したがって,偶数の不足数も無数に存在する.

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