11330001119993 has 2 divisors, whose sum is σ = 11330001119994. Its totient is φ = 11330001119992.

The previous prime is 11330001119959. The next prime is 11330001120013. The reversal of 11330001119993 is 39991110003311.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 11326671022144 + 3330097849 = 3365512^2 + 57707^2 .

It is an emirp because it is prime and its reverse (39991110003311) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 11330001119993 - 2⁴⁰ = 10230489492217 is a prime.

It is a super-3 number, since 3×11330001119993³ (a number of 40 digits) contains 333 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 11330001119993.

It is not a weakly prime, because it can be changed into another prime (11330001119893) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5665000559996 + 5665000559997.

It is an arithmetic number, because the mean of its divisors is an integer number (5665000559997).

Almost surely, 2^{11330001119993} is an apocalyptic number.

It is an amenable number.

11330001119993 is a deficient number, since it is larger than the sum of its proper divisors (1).

11330001119993 is an equidigital number, since it uses as much as digits as its factorization.

11330001119993 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 19683, while the sum is 41.

The spelling of 11330001119993 in words is "eleven trillion, three hundred thirty billion, one million, one hundred nineteen thousand, nine hundred ninety-three".