10100111012113 has 4 divisors (see below), whose sum is σ = 10109721014128. Its totient is φ = 10090501010100.

The previous prime is 10100111012087. The next prime is 10100111012141. The reversal of 10100111012113 is 31121011100101.

10100111012113 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

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-10100111012113 is a prime.

It is a super-4 number, since 4×10100111012113⁴ (a number of 53 digits) contains 4444 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 10100111012093 and 10100111012102.

It is not an unprimeable number, because it can be changed into a prime (10100111012153) 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, 4804999431 + ... + 4805001532.

It is an arithmetic number, because the mean of its divisors is an integer number (2527430253532).

Almost surely, 2^{10100111012113} is an apocalyptic number.

It is an amenable number.

10100111012113 is a deficient number, since it is larger than the sum of its proper divisors (9610002015).

10100111012113 is an equidigital number, since it uses as much as digits as its factorization.

10100111012113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 9610002014.

The product of its (nonzero) digits is 6, while the sum is 13.

Adding to 10100111012113 its reverse (31121011100101), we get a palindrome (41221122112214).

The spelling of 10100111012113 in words is "ten trillion, one hundred billion, one hundred eleven million, twelve thousand, one hundred thirteen".