12114214331 has 4 divisors (see below), whose sum is σ = 12826815192. Its totient is φ = 11401613472.

The previous prime is 12114214303. The next prime is 12114214361. The reversal of 12114214331 is 13341241121.

12114214331 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 an emirpimes, since its reverse is a distinct semiprime: 13341241121 = 13 ⋅1026249317.

It is a cyclic number.

It is not a de Polignac number, because 12114214331 - 2⁶ = 12114214267 is a prime.

It is a super-3 number, since 3×12114214331³ (a number of 31 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 12114214297 and 12114214306.

It is not an unprimeable number, because it can be changed into a prime (12114214361) 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, 356300405 + ... + 356300438.

It is an arithmetic number, because the mean of its divisors is an integer number (3206703798).

Almost surely, 2^12114214331 is an apocalyptic number.

12114214331 is a deficient number, since it is larger than the sum of its proper divisors (712600861).

12114214331 is an equidigital number, since it uses as much as digits as its factorization.

12114214331 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 712600860.

The product of its digits is 576, while the sum is 23.

Adding to 12114214331 its reverse (13341241121), we get a palindrome (25455455452).

The spelling of 12114214331 in words is "twelve billion, one hundred fourteen million, two hundred fourteen thousand, three hundred thirty-one".