13104032024113 has 4 divisors (see below), whose sum is σ = 13104046984048. Its totient is φ = 13104017064180.

The previous prime is 13104032024071. The next prime is 13104032024159. The reversal of 13104032024113 is 31142023040131.

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 a de Polignac number, because none of the positive numbers 2^k-13104032024113 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (13104032024173) 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, 6078531 + ... + 7947112.

It is an arithmetic number, because the mean of its divisors is an integer number (3276011746012).

Almost surely, 2^{13104032024113} is an apocalyptic number.

It is an amenable number.

13104032024113 is a deficient number, since it is larger than the sum of its proper divisors (14959935).

13104032024113 is an equidigital number, since it uses as much as digits as its factorization.

13104032024113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 14959934.

The product of its (nonzero) digits is 1728, while the sum is 25.

Adding to 13104032024113 its reverse (31142023040131), we get a palindrome (44246055064244).

The spelling of 13104032024113 in words is "thirteen trillion, one hundred four billion, thirty-two million, twenty-four thousand, one hundred thirteen".