11003120113 has 4 divisors (see below), whose sum is σ = 11006414848. Its totient is φ = 10999825380.

The previous prime is 11003120087. The next prime is 11003120119. The reversal of 11003120113 is 31102130011.

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: 31102130011 = 5387 ⋅5773553.

It is a cyclic number.

It is not a de Polignac number, because 11003120113 - 2²¹ = 11001022961 is a prime.

It is a super-2 number, since 2×11003120113² (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 = 11003120093 and 11003120102.

It is not an unprimeable number, because it can be changed into a prime (11003120119) 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, 1642353 + ... + 1649038.

It is an arithmetic number, because the mean of its divisors is an integer number (2751603712).

Almost surely, 2^11003120113 is an apocalyptic number.

It is an amenable number.

11003120113 is a deficient number, since it is larger than the sum of its proper divisors (3294735).

11003120113 is an equidigital number, since it uses as much as digits as its factorization.

11003120113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 3294734.

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

Adding to 11003120113 its reverse (31102130011), we get a palindrome (42105250124).

It can be divided in two parts, 1100 and 3120113, that added together give a palindrome (3121213).

The spelling of 11003120113 in words is "eleven billion, three million, one hundred twenty thousand, one hundred thirteen".