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Playing with divisors
@hari_2425 You are using a wrong approach.
Number of divisor of number n,x= product(ai+1) (ai is the power of primes).
Now with a bit of maths you will get product of all div= n^x (which is our p)
now you need to find power of primes in this new number.
again with bit of maths you will get that the new power of ith prime number is (ai*P)/2
Now again you have all the power calculated using the above methods, use the formula to cal number of divisors.
Just one thing as we are doing modulo operations keep in mind the inverse modulo of 2(which you have done buy have not taken care of P).
If this resolves your doubt mark its as resolved.
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