0
$\begingroup$

So by Fundamental Arithmetic Theorem, any integer has a unique prime factorization into primes, written as: $$n=p_1^{k_1}p_2^{k_2}p_3^{k_3}...p_r^{k_r}$$ From exponents $k_1,...k_r$ it is possible to find the number of divisors of $n$, called divisor function written as $d(n)$. Find $d(100)$ and $d(320)$. Then, find a sgeneral formula for $d(n)$ in terms of $k_1,...,k_r.$

Having trouble figuring out where to start here. Any guidance would be really great! Thank you.

$\endgroup$
3
$\begingroup$

Hint

A divisor of $n$ can be written (uniquely) in the form $$ n=p_{1}^{\alpha_1}p_{2}^{\alpha_2}\dotsb p_{r}^{\alpha_r} $$ where $0\leq \alpha_i\leq k_{i}$. How many choices are there for $\alpha_1$, for $\alpha_2$, and so on? Then use the multiplication principle to get the result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.