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Evaluate $ \sigma(210)$ and $d(63)$

I'm not sure if I got this correct, so here is my attempt.

By Theorem 6.3, suppose we have $n=p_1^{\alpha 1}...p_s^{\alpha s}$, then $d(n) =(\alpha_1 +1)(\alpha_2 +1)...(\alpha_s +1)$ and $\sigma(n) =\frac{p_1^{\alpha_1+1}-1}{p_1-1}\frac{p_2^{\alpha_2+1}-1}{p_2-1}...\frac{p_1^{\alpha_s+1}-1}{p_s-1}$

For $ \sigma(210)$ we have $ \sigma(3 \cdot 5 \cdot 2 \cdot 7)$

$\sigma(210) =\frac{3^{1+1}-1}{3-1} \cdot \frac{5^{1+1}-1}{5-1} \cdot \frac{2^{1+1}-1}{2-1} \cdot \frac{7^{1+1}-1}{7-1}$

$\sigma(210) =\frac{3^{2}-1}{3-1} \cdot \frac{5^{2}-1}{5-1} \cdot \frac{2^{2}-1}{2-1} \cdot \frac{7^{2}-1}{7-1}$

$\sigma(210) =\frac{9-1}{3-1} \cdot \frac{25-1}{5-1} \cdot \frac{4-1}{2-1} \cdot \frac{49-1}{7-1}$

$\sigma(210) =\frac{8}{2} \cdot \frac{24}{4} \cdot \frac{3}{1} \cdot \frac{48}{6}$

$\sigma(210) = 4 \cdot 6 \cdot 3 \cdot 8 \rightarrow 576$

I forgot to mention that whenever we have a $p$ which denotes a prime, the only divisors are $1$ and $p$, so $d(p) =2$ and $\sigma(p) = p+1$. Moreover, $d(p^n) = n+1$ occurs whenever we have positive divisors such as $1,p,p^2,...p^n$

For $d(63)$, we have

$d(63) \rightarrow d(7 \cdot 3^2)$

$d(7)d(9)$

Since $7$ is a prime we have $d(7) =2$ and for $d(9)$ we have $d(3^2) \rightarrow d(3^2)=2+1 =3$. Therefore, the answer is $(2)(3) = 6$

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  • $\begingroup$ Your answers are right. For $\sigma(2\cdot 3\cdot 5\cdot 7$, it is easier to just write $(1+2)(1+3)(1+5)(1+7)$ instead of the one that comes from summing geometric series. $\endgroup$ Oct 21, 2014 at 1:25
  • $\begingroup$ why is it easier to write (1+2)(1+3)(1+5)(1+7)? $\endgroup$
    – usukidoll
    Oct 21, 2014 at 1:46
  • $\begingroup$ At least purely in terms of typing, your $\sigma(210) =\frac{3^{2}-1}{3-1} \cdot \frac{5^{2}-1}{5-1} \cdot \frac{2^{2}-1}{2-1} \cdot \frac{7^{2}-1}{7-1}$ is more work. Makes something simple look more complicated than it is. $\endgroup$ Oct 21, 2014 at 1:50
  • $\begingroup$ But if I do it that way, I sort of understand it better. It's what Number Theory by George Andrews did. $\endgroup$
    – usukidoll
    Oct 21, 2014 at 1:50
  • $\begingroup$ If it is more helpful to you, fine. In teaching number theory, I have found that a common problem students have in proof-making is reliance on formulas. $\endgroup$ Oct 21, 2014 at 1:54

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