12143331113 has 4 divisors (see below), whose sum is σ = 12143751168. Its totient is φ = 12142911060.

The previous prime is 12143331059. The next prime is 12143331119. The reversal of 12143331113 is 31113334121.

12143331113 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 12143331113 - 2¹² = 12143327017 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (12143331119) 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, 163181 + ... + 225642.

It is an arithmetic number, because the mean of its divisors is an integer number (3035937792).

Almost surely, 2^12143331113 is an apocalyptic number.

It is an amenable number.

12143331113 is a deficient number, since it is larger than the sum of its proper divisors (420055).

12143331113 is an equidigital number, since it uses as much as digits as its factorization.

12143331113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 420054.

The product of its digits is 648, while the sum is 23.

Adding to 12143331113 its reverse (31113334121), we get a palindrome (43256665234).

The spelling of 12143331113 in words is "twelve billion, one hundred forty-three million, three hundred thirty-one thousand, one hundred thirteen".