13000100210113 has 4 divisors (see below), whose sum is σ = 13002348201408. Its totient is φ = 12997852218820.

The previous prime is 13000100210099. The next prime is 13000100210159. The reversal of 13000100210113 is 31101200100031.

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 13000100210113 - 2⁹ = 13000100209601 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 13000100210093 and 13000100210102.

It is not an unprimeable number, because it can be changed into a prime (13000170210113) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1123986973 + ... + 1123998538.

It is an arithmetic number, because the mean of its divisors is an integer number (3250587050352).

Almost surely, 2^{13000100210113} is an apocalyptic number.

It is an amenable number.

13000100210113 is a deficient number, since it is larger than the sum of its proper divisors (2247991295).

13000100210113 is an equidigital number, since it uses as much as digits as its factorization.

13000100210113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 2247991294.

The product of its (nonzero) digits is 18, while the sum is 13.

Adding to 13000100210113 its reverse (31101200100031), we get a palindrome (44101300310144).

The spelling of 13000100210113 in words is "thirteen trillion, one hundred million, two hundred ten thousand, one hundred thirteen".