12022143433 has 4 divisors (see below), whose sum is σ = 12301728208. Its totient is φ = 11742558660.

The previous prime is 12022143409. The next prime is 12022143439. The reversal of 12022143433 is 33434122021.

12022143433 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

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 12022143433 - 2¹³ = 12022135241 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 12022143398 and 12022143407.

It is not an unprimeable number, because it can be changed into a prime (12022143439) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 139792323 + ... + 139792408.

It is an arithmetic number, because the mean of its divisors is an integer number (3075432052).

Almost surely, 2^12022143433 is an apocalyptic number.

It is an amenable number.

12022143433 is a deficient number, since it is larger than the sum of its proper divisors (279584775).

12022143433 is an equidigital number, since it uses as much as digits as its factorization.

12022143433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 279584774.

The product of its (nonzero) digits is 3456, while the sum is 25.

Adding to 12022143433 its reverse (33434122021), we get a palindrome (45456265454).

The spelling of 12022143433 in words is "twelve billion, twenty-two million, one hundred forty-three thousand, four hundred thirty-three".