23124000113 has 4 divisors (see below), whose sum is σ = 23402602608. Its totient is φ = 22845397620.

The previous prime is 23124000101. The next prime is 23124000121. The reversal of 23124000113 is 31100042132.

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 23124000113 - 2¹⁰ = 23123999089 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 23124000091 and 23124000100.

It is not an unprimeable number, because it can be changed into a prime (23124000143) 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, 139301123 + ... + 139301288.

It is an arithmetic number, because the mean of its divisors is an integer number (5850650652).

Almost surely, 2^23124000113 is an apocalyptic number.

It is an amenable number.

23124000113 is a deficient number, since it is larger than the sum of its proper divisors (278602495).

23124000113 is an equidigital number, since it uses as much as digits as its factorization.

23124000113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 278602494.

The product of its (nonzero) digits is 144, while the sum is 17.

Adding to 23124000113 its reverse (31100042132), we get a palindrome (54224042245).

The spelling of 23124000113 in words is "twenty-three billion, one hundred twenty-four million, one hundred thirteen", and thus it is an aban number.