11313000403 has 4 divisors (see below), whose sum is σ = 11498459488. Its totient is φ = 11127541320.

The previous prime is 11313000401. The next prime is 11313000497. The reversal of 11313000403 is 30400031311.

It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 30400031311 = 13 ⋅2338463947.

It is a cyclic number.

It is not a de Polignac number, because 11313000403 - 2¹ = 11313000401 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (11313000401) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 92729451 + ... + 92729572.

It is an arithmetic number, because the mean of its divisors is an integer number (2874614872).

Almost surely, 2^11313000403 is an apocalyptic number.

11313000403 is a deficient number, since it is larger than the sum of its proper divisors (185459085).

11313000403 is an equidigital number, since it uses as much as digits as its factorization.

11313000403 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 185459084.

The product of its (nonzero) digits is 108, while the sum is 16.

Adding to 11313000403 its reverse (30400031311), we get a palindrome (41713031714).

The spelling of 11313000403 in words is "eleven billion, three hundred thirteen million, four hundred three", and thus it is an aban number.