11983412707991 has 2 divisors, whose sum is σ = 11983412707992. Its totient is φ = 11983412707990.

The previous prime is 11983412707979. The next prime is 11983412707993. The reversal of 11983412707991 is 19970721438911.

11983412707991 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-11983412707991 is a prime.

Together with 11983412707993, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (11983412707993) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5991706353995 + 5991706353996.

It is an arithmetic number, because the mean of its divisors is an integer number (5991706353996).

Almost surely, 2^{11983412707991} is an apocalyptic number.

11983412707991 is a deficient number, since it is larger than the sum of its proper divisors (1).

11983412707991 is an equidigital number, since it uses as much as digits as its factorization.

11983412707991 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 6858432, while the sum is 62.

The spelling of 11983412707991 in words is "eleven trillion, nine hundred eighty-three billion, four hundred twelve million, seven hundred seven thousand, nine hundred ninety-one".