10113003030113 has 4 divisors (see below), whose sum is σ = 10192632975360. Its totient is φ = 10033373084868.

The previous prime is 10113003030101. The next prime is 10113003030161. The reversal of 10113003030113 is 31103030031101.

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-10113003030113 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 10113003030091 and 10113003030100.

It is not an unprimeable number, because it can be changed into a prime (10113003030313) 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, 39814972433 + ... + 39814972686.

It is an arithmetic number, because the mean of its divisors is an integer number (2548158243840).

Almost surely, 2^{10113003030113} is an apocalyptic number.

It is an amenable number.

10113003030113 is a deficient number, since it is larger than the sum of its proper divisors (79629945247).

10113003030113 is an equidigital number, since it uses as much as digits as its factorization.

10113003030113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 79629945246.

The product of its (nonzero) digits is 81, while the sum is 17.

Adding to 10113003030113 its reverse (31103030031101), we get a palindrome (41216033061214).

The spelling of 10113003030113 in words is "ten trillion, one hundred thirteen billion, three million, thirty thousand, one hundred thirteen".