12110314433 has 4 divisors (see below), whose sum is σ = 13840359360. Its totient is φ = 10380269508.

The previous prime is 12110314423. The next prime is 12110314457. The reversal of 12110314433 is 33441301121.

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 not a de Polignac number, because 12110314433 - 2¹⁶ = 12110248897 is a prime.

It is a super-3 number, since 3×12110314433³ (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 = 12110314399 and 12110314408.

It is not an unprimeable number, because it can be changed into a prime (12110314423) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 865022453 + ... + 865022466.

It is an arithmetic number, because the mean of its divisors is an integer number (3460089840).

Almost surely, 2^12110314433 is an apocalyptic number.

It is an amenable number.

12110314433 is a deficient number, since it is larger than the sum of its proper divisors (1730044927).

12110314433 is an equidigital number, since it uses as much as digits as its factorization.

12110314433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1730044926.

The product of its (nonzero) digits is 864, while the sum is 23.

Adding to 12110314433 its reverse (33441301121), we get a palindrome (45551615554).

The spelling of 12110314433 in words is "twelve billion, one hundred ten million, three hundred fourteen thousand, four hundred thirty-three".