10040131113 has 4 divisors (see below), whose sum is σ = 13386841488. Its totient is φ = 6693420740.

The previous prime is 10040131091. The next prime is 10040131141. The reversal of 10040131113 is 31113104001.

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 10040131113 - 2⁵ = 10040131081 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 10040131092 and 10040131101.

It is not an unprimeable number, because it can be changed into a prime (10040131613) 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, 1673355183 + ... + 1673355188.

It is an arithmetic number, because the mean of its divisors is an integer number (3346710372).

Almost surely, 2^10040131113 is an apocalyptic number.

It is an amenable number.

10040131113 is a deficient number, since it is larger than the sum of its proper divisors (3346710375).

10040131113 is an equidigital number, since it uses as much as digits as its factorization.

10040131113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 3346710374.

The product of its (nonzero) digits is 36, while the sum is 15.

Adding to 10040131113 its reverse (31113104001), we get a palindrome (41153235114).

The spelling of 10040131113 in words is "ten billion, forty million, one hundred thirty-one thousand, one hundred thirteen".