11983412708193 has 4 divisors (see below), whose sum is σ = 15977883610928. Its totient is φ = 7988941805460.

The previous prime is 11983412708191. The next prime is 11983412708213. The reversal of 11983412708193 is 39180721438911.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is not a de Polignac number, because 11983412708193 - 2¹ = 11983412708191 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (11983412708191) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1997235451363 + ... + 1997235451368.

It is an arithmetic number, because the mean of its divisors is an integer number (3994470902732).

Almost surely, 2^{11983412708193} is an apocalyptic number.

It is an amenable number.

11983412708193 is a deficient number, since it is larger than the sum of its proper divisors (3994470902735).

11983412708193 is an equidigital number, since it uses as much as digits as its factorization.

11983412708193 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 3994470902734.

The product of its (nonzero) digits is 2612736, while the sum is 57.

The spelling of 11983412708193 in words is "eleven trillion, nine hundred eighty-three billion, four hundred twelve million, seven hundred eight thousand, one hundred ninety-three".