【数学A】例題1.1.7：約数の個数・総和（One More）★★

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% 例題A1.1.7：約数の個数・総和（One More）★★
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72の正の約数の個数とその総和を求めよ．

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% 例題A1.1.7の解答（One More）★★
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72を素因数分解すると，$72=2^3\times 3^2$

\[
(3+1)\times(2+1)=12
\]

より，約数の個数は，12個

また，約数の総和は，

\[
(1+2+2^2+2^3)(1+3+3^2)=15\times 13=195
\]

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% 問題A1.1.7：約数の個数・総和（One More）★★
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120の正の約数の個数とその総和を求めよ．

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% 問題A1.1.7の解答（One More）★★
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120を素因数分解すると，$120=2^3\times 3\times 5$

\[
(3+1)\times(1+1)\times(1+1)=16
\]

より，約数の個数は，16個

また，約数の総和は，

\[
(1+2+2^2+2^3)(1+3)(1+5)=15\times 4\times 6=360
\]

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% 例題A1.1.7の側注：約数の個数・総和（One More）★★
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% 側注1
積の法則を用いる．なお，「$3\times 2=6$より，約数の個数は，6個」などと答えてしまわないように注意すること．

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% 問題A1.1.7の側注：約数の個数・総和（One More）★★
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% 側注1
積の法則を用いる．

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% 例題A1.1.7の考え方・補足など：約数の個数・総和（One More）★★
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% 考え方
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72を素因数分解すると，$72=2^{3}\times 3^{2}$であるから，正の約数は，次のようになる．

\[
1\times 1,2\times 1,2^2\times 1,2^3\times 1,1\times 3,2\times 3,2^2\times 3,2^3\times 3,1\times 3^2,2\times 3^2,2^2\times 3^2,2^3\times 3^2\cdots(\mathrm{i})
\]

72の正の約数は，因数2を0，1，2，3個（4通り）のいずれかをもち，因数3を0，1，2個（3通り）のいずれかをもつことから，積の法則より，
\[
4\times 3=12\text{(個)}
\]

また，$(1+2+2^2+2^3)(1+3+3^2)$を展開すると，

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

これは，(i)の総和と一致し，約数の総和は$(1+2+2^2+2^3)(1+3+3^2)$で求めることができる．

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% 正の約数の個数と総和
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自然数$N$が$N=p^aq^br^c\cdots$と素因数分解されているとき，

正の約数の個数は，

\[
(a+1)(b+1)(c+1)\cdots(個)
\]

正の約数の総和は，

\[
(1+p+p^2+\cdots+p^a)(1+q+q^2+\cdots+q^b)(1+r+r^2+\cdots+r^c)\cdots
\]