Let "n= 2^{31}3^{19}. How many positive divisors of n^2 are less than n but
do not divide n?
the answer is 589
please give me the complete solution....
thanks in advance
"Raise" is an odd notation! Powers are commonly written as "^". I think you mean "n= 2^{31}3^{19}" or, better, using LaTex, "$\displaystyle n= 2^{31}3^{19}$". $\displaystyle n^2$ then is $\displaystyle n^2= 2^{62}3^{38}$. To be a divisor of that but not of n itself, a number must be divisible by either $\displaystyle 2^{32}$ or $\displaystyle 3^{20}$.
Now, what additional factors can you add and still keep the number less than n?