11983412708947 has 2 divisors, whose sum is σ = 11983412708948. Its totient is φ = 11983412708946.

The previous prime is 11983412708923. The next prime is 11983412709011. The reversal of 11983412708947 is 74980721438911.

11983412708947 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is an a-pointer prime, because the next prime (11983412709011) can be obtained adding 11983412708947 to its sum of digits (64).

It is a weak prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-11983412708947 is a prime.

It is not a weakly prime, because it can be changed into another prime (11983412708747) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5991706354473 + 5991706354474.

It is an arithmetic number, because the mean of its divisors is an integer number (5991706354474).

Almost surely, 2^{11983412708947} is an apocalyptic number.

11983412708947 is a deficient number, since it is larger than the sum of its proper divisors (1).

11983412708947 is an equidigital number, since it uses as much as digits as its factorization.

11983412708947 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 24385536, while the sum is 64.

The spelling of 11983412708947 in words is "eleven trillion, nine hundred eighty-three billion, four hundred twelve million, seven hundred eight thousand, nine hundred forty-seven".