Arithmetic

Factors

given number $n$
represent it in prime factorization $a^p b^q c^r...$
so, $a$ can acquire powers $\{0, 1, 2, ..., p\}$, $n(\{0, 1, 2, ..., p\}) = p+1$
so, $b$ can acquire powers $\{0, 1, 2, ..., q\}$, $n(\{0, 1, 2, ..., p\}) = q+1$
so, $c$ can acquire powers $\{0, 1, 2, ..., r\}$, $n(\{0, 1, 2, ..., p\}) = r+1$
and so on, here $n(\text{A})$ is to represent the cardinality of set $\text{A}$
so total no. of factors are $(p+1)(q+1)(r+1)$

should have at least one 2
given number $n$
represent it in prime factorization $2^p b^q c^r...$
so, $a$ can acquire powers $\{1, 2, ..., p\}$, $n(\{1, 2, ..., p\}) = p$
so, $b$ can acquire powers $\{0, 1, 2, ..., q\}$, $n(\{0, 1, 2, ..., p\}) = q+1$
so, $c$ can acquire powers $\{0, 1, 2, ..., r\}$, $n(\{0, 1, 2, ..., p\}) = r+1$
and so on, here $n(\text{A})$ is to represent the cardinality of set $\text{A}$
so total no. of even factors are $(p)(q+1)(r+1)$

given number $n$
represent it in prime factorization $a^p b^q c^r...$
so, $a$ can acquire powers $\{0, 2, 4, ..., p\}$, $n(\{0, 1, 2, ..., p\}) = \lceil{\frac{p+1}{2}}\rceil$
so, $b$ can acquire powers $\{0, 2, 4, ..., q\}$, $n(\{0, 1, 2, ..., p\}) = \lceil{\frac{q+1}{2}}\rceil$
so, $c$ can acquire powers $\{0, 2, 4, ..., r\}$, $n(\{0, 1, 2, ..., p\}) = \lceil{\frac{r+1}{2}}\rceil$
and so on, here $n(\text{A})$ is to represent the cardinality of set $\text{A}$
so total no. of factors are

given, $n$ and its prime factorization $a^p b^q c^r...$
$f(n) = (p+1)(q+1)(r+1)...$
$F(n) = \frac{1}{2}f(n)$,
- is not possible when $f(n)$ is odd
- if, $p, q, r,...$ are all even then $n$ is a prefect square
  - that means that $\sqrt{n}$ is a whole no.
  - therefore $F(n) = \frac{1}{2}f(n) - 1$, $\sqrt{n} \times \sqrt{n}$ is not counted

\[\frac{a^{p+1} - 1}{a-1} \frac{b^{q+1} - 1}{b-1} \frac{c^{r+1} - 1}{c-1} ...\]

co-primes - there gcd (hcf) is 1
given, $n$ and its prime factorization $a^p b^q c^r...$
$n(\{a, b, c, ...\}) = x$, i.e. no of unique prime in prime factorization is $x$
then, total no of co-primes are $2^{x-1}$

given, $n$ and its prime factorization $a^p b^q c^r...$
then, $$ \phi(n) = n\Big(1-\frac{1}{a}\Big) \Big(1-\frac{1}{b} \Big) \Big(1-\frac{1}{c}\Big) $$
so number of coprimes to $n$ that are less than $n$ are:
- $\frac{n}{2} \times \phi(n)$

\[ \left\lceil\frac{p+1}{2}\right\rceil \left\lceil\frac{q+1}{2}\right\rceil \left\lceil\frac{r+1}{2}\right\rceil \]

$n = a^p b^q c^r...$
$m = a^l b^m c^n...$
total no of common factors are $(\min(p, l) + 1) \times (\min(q, m) + 1) \times (\min(r, m) + 1) ...$