12114312101 has 2 divisors, whose sum is σ = 12114312102. Its totient is φ = 12114312100.

The previous prime is 12114312073. The next prime is 12114312103. The reversal of 12114312101 is 10121341121.

12114312101 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 6851535076 + 5262777025 = 82774^2 + 72545^2 .

It is a cyclic number.

It is not a de Polignac number, because 12114312101 - 2¹⁴ = 12114295717 is a prime.

It is a super-3 number, since 3×12114312101³ (a number of 31 digits) contains 333 as substring.

Together with 12114312103, it forms a pair of twin primes.

It is a Chen prime.

It is a self number, because there is not a number n which added to its sum of digits gives 12114312101.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (12114312103) 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 as a sum of consecutive naturals, namely, 6057156050 + 6057156051.

It is an arithmetic number, because the mean of its divisors is an integer number (6057156051).

Almost surely, 2^12114312101 is an apocalyptic number.

It is an amenable number.

12114312101 is a deficient number, since it is larger than the sum of its proper divisors (1).

12114312101 is an equidigital number, since it uses as much as digits as its factorization.

12114312101 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 48, while the sum is 17.

Adding to 12114312101 its reverse (10121341121), we get a palindrome (22235653222).

The spelling of 12114312101 in words is "twelve billion, one hundred fourteen million, three hundred twelve thousand, one hundred one".