13100120113 has 4 divisors (see below), whose sum is σ = 13100590080. Its totient is φ = 13099650148.

The previous prime is 13100120101. The next prime is 13100120123. The reversal of 13100120113 is 31102100131.

13100120113 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

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: 31102100131 = 107 ⋅290673833.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-13100120113 is a prime.

It is a super-2 number, since 2×13100120113² (a number of 21 digits) contains 22 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 13100120093 and 13100120102.

It is not an unprimeable number, because it can be changed into a prime (13100120123) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 190345 + ... + 249862.

It is an arithmetic number, because the mean of its divisors is an integer number (3275147520).

Almost surely, 2^13100120113 is an apocalyptic number.

It is an amenable number.

13100120113 is a deficient number, since it is larger than the sum of its proper divisors (469967).

13100120113 is an equidigital number, since it uses as much as digits as its factorization.

13100120113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 469966.

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

Adding to 13100120113 its reverse (31102100131), we get a palindrome (44202220244).

The spelling of 13100120113 in words is "thirteen billion, one hundred million, one hundred twenty thousand, one hundred thirteen".