10123010000113 has 4 divisors (see below), whose sum is σ = 10155559871808. Its totient is φ = 10090460128420.

The previous prime is 10123010000039. The next prime is 10123010000119. The reversal of 10123010000113 is 31100001032101.

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 10123010000113 - 2³³ = 10114420065521 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 10123010000093 and 10123010000102.

It is not an unprimeable number, because it can be changed into a prime (10123010000119) 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, 16274935381 + ... + 16274936002.

It is an arithmetic number, because the mean of its divisors is an integer number (2538889967952).

Almost surely, 2^{10123010000113} is an apocalyptic number.

It is an amenable number.

10123010000113 is a deficient number, since it is larger than the sum of its proper divisors (32549871695).

10123010000113 is an equidigital number, since it uses as much as digits as its factorization.

10123010000113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 32549871694.

The product of its (nonzero) digits is 18, while the sum is 13.

Adding to 10123010000113 its reverse (31100001032101), we get a palindrome (41223011032214).

The spelling of 10123010000113 in words is "ten trillion, one hundred twenty-three billion, ten million, one hundred thirteen", and thus it is an aban number.