10100000113 has 4 divisors (see below), whose sum is σ = 11542857280. Its totient is φ = 8657142948.

The previous prime is 10100000111. The next prime is 10100000129. The reversal of 10100000113 is 31100000101.

It is a happy number.

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, and also an emirpimes, since its reverse is a distinct semiprime: 31100000101 = 44021 ⋅706481.

It is a cyclic number.

It is not a de Polignac number, because 10100000113 - 2¹ = 10100000111 is a prime.

It is a Harshad number since it is a multiple of its sum of digits (7), and also a Moran number because the ratio is a prime number: 1442857159 = 10100000113 / (1 + 0 + 1 + 0 + 0 + 0 + 0 + 0 + 1 + 1 + 3).

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 10100000096 and 10100000105.

It is not an unprimeable number, because it can be changed into a prime (10100000111) 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, 721428573 + ... + 721428586.

It is an arithmetic number, because the mean of its divisors is an integer number (2885714320).

Almost surely, 2^10100000113 is an apocalyptic number.

It is an amenable number.

10100000113 is a deficient number, since it is larger than the sum of its proper divisors (1442857167).

10100000113 is an equidigital number, since it uses as much as digits as its factorization.

10100000113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 1442857166.

The product of its (nonzero) digits is 3, while the sum is 7.

Adding to 10100000113 its reverse (31100000101), we get a palindrome (41200000214).

The spelling of 10100000113 in words is "ten billion, one hundred million, one hundred thirteen", and thus it is an aban number.