12114331433 has 4 divisors (see below), whose sum is σ = 12133231188. Its totient is φ = 12095431680.

The previous prime is 12114331421. The next prime is 12114331517. The reversal of 12114331433 is 33413341121.

It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 33413341121 = 152213 ⋅219517.

It can be written as a sum of positive squares in 2 ways, for example, as 11765090089 + 349241344 = 108467^2 + 18688^2 .

It is a cyclic number.

It is not a de Polignac number, because 12114331433 - 2¹⁰ = 12114330409 is a prime.

It is a super-3 number, since 3×12114331433³ (a number of 31 digits) contains 333 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (12114330433) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 9448916 + ... + 9450197.

It is an arithmetic number, because the mean of its divisors is an integer number (3033307797).

Almost surely, 2^12114331433 is an apocalyptic number.

It is an amenable number.

12114331433 is a deficient number, since it is larger than the sum of its proper divisors (18899755).

12114331433 is an equidigital number, since it uses as much as digits as its factorization.

12114331433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 18899754.

The product of its digits is 2592, while the sum is 26.

Adding to 12114331433 its reverse (33413341121), we get a palindrome (45527672554).

The spelling of 12114331433 in words is "twelve billion, one hundred fourteen million, three hundred thirty-one thousand, four hundred thirty-three".