12003113133001 has 4 divisors (see below), whose sum is σ = 12003567840000. Its totient is φ = 12002658426004.

The previous prime is 12003113132999. The next prime is 12003113133043. The reversal of 12003113133001 is 10033131130021.

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 a cyclic number.

It is not a de Polignac number, because 12003113133001 - 2¹ = 12003113132999 is a prime.

It is a Duffinian number.

It is a self number, because there is not a number n which added to its sum of digits gives 12003113133001.

It is not an unprimeable number, because it can be changed into a prime (12003113133091) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 227313901 + ... + 227366698.

It is an arithmetic number, because the mean of its divisors is an integer number (3000891960000).

Almost surely, 2^{12003113133001} is an apocalyptic number.

It is an amenable number.

12003113133001 is a deficient number, since it is larger than the sum of its proper divisors (454706999).

12003113133001 is an equidigital number, since it uses as much as digits as its factorization.

12003113133001 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 454706998.

The product of its (nonzero) digits is 162, while the sum is 19.

Adding to 12003113133001 its reverse (10033131130021), we get a palindrome (22036244263022).

The spelling of 12003113133001 in words is "twelve trillion, three billion, one hundred thirteen million, one hundred thirty-three thousand, one".