Thread: difficult problem

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, " ". then is . To be a divisor of that but not of n itself, a number must be divisible by either or .

Now, what additional factors can you add and still keep the number less than n?