Question

HELP with #11

http://prntscr.com/4feqfk

Answer 1

\(\large \tau \) : number of divisors
\(\large \sigma \) : sim of divisors

Answer 2

we might could use
\(\large 1<\frac{\sigma(n)}{n}<\prod \frac{pi}{pi-1} \)

Answer 3

for the second part

Answer 4

yeah have u worked \(\tau(n)\) already ?

Answer 5

\(\tau (n) = \sum_{d|n} 1=k \)

so xD there is infinitly many compination we ould get 1+1+1+1.. =k
it dont need work , its common sense , unlike second part >.<
i think the question mintioned both to show the diffrence