12013403533 has 4 divisors (see below), whose sum is σ = 12015471040.
Its totient is φ = 12011336028.
The previous prime is 12013403513. The next prime is 12013403609. The reversal of 12013403533 is 33530431021.
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, and also an emirpimes, since its reverse is a distinct semiprime: 33530431021 = 23 ⋅1457844827.
It is a cyclic number.
It is not a de Polignac number, because 12013403533 - 229 = 11476532621 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 12013403498 and 12013403507.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (12013403513) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1025013 + ... + 1036666.
It is an arithmetic number, because the mean of its divisors is an integer number (3003867760).
Almost surely, 212013403533 is an apocalyptic number.
It is an amenable number.
12013403533 is a deficient number, since it is larger than the sum of its proper divisors (2067507).
12013403533 is an equidigital number, since it uses as much as digits as its factorization.
12013403533 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 2067506.
The product of its (nonzero) digits is 3240, while the sum is 25.
Adding to 12013403533 its reverse (33530431021), we get a palindrome (45543834554).
The spelling of 12013403533 in words is "twelve billion, thirteen million, four hundred three thousand, five hundred thirty-three".