12711344163853 has 4 divisors (see below), whose sum is σ = 13380362277760. Its totient is φ = 12042326049948.

The previous prime is 12711344163841. The next prime is 12711344163883. The reversal of 12711344163853 is 35836144311721.

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 12711344163853 - 2²⁵ = 12711310609421 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 12711344163797 and 12711344163806.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (12711344163883) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 334509056925 + ... + 334509056962.

It is an arithmetic number, because the mean of its divisors is an integer number (3345090569440).

Almost surely, 2^{12711344163853} is an apocalyptic number.

It is an amenable number.

12711344163853 is a deficient number, since it is larger than the sum of its proper divisors (669018113907).

12711344163853 is an equidigital number, since it uses as much as digits as its factorization.

12711344163853 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 669018113906.

The product of its digits is 1451520, while the sum is 49.

The spelling of 12711344163853 in words is "twelve trillion, seven hundred eleven billion, three hundred forty-four million, one hundred sixty-three thousand, eight hundred fifty-three".