12114343303 has 4 divisors (see below), whose sum is σ = 12175219600. Its totient is φ = 12053467008.

The previous prime is 12114343301. The next prime is 12114343321. The reversal of 12114343303 is 30334341121.

It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 30334341121 = 7 ⋅4333477303.

It is a cyclic number.

It is not a de Polignac number, because 12114343303 - 2¹ = 12114343301 is a prime.

It is a super-3 number, since 3×12114343303³ (a number of 31 digits) contains 333 as substring.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (12114343301) 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, 30437850 + ... + 30438247.

It is an arithmetic number, because the mean of its divisors is an integer number (3043804900).

Almost surely, 2^12114343303 is an apocalyptic number.

12114343303 is a deficient number, since it is larger than the sum of its proper divisors (60876297).

12114343303 is an equidigital number, since it uses as much as digits as its factorization.

12114343303 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 60876296.

The product of its (nonzero) digits is 2592, while the sum is 25.

Adding to 12114343303 its reverse (30334341121), we get a palindrome (42448684424).

The spelling of 12114343303 in words is "twelve billion, one hundred fourteen million, three hundred forty-three thousand, three hundred three".