11000113 has 2 divisors, whose sum is σ = 11000114. Its totient is φ = 11000112.

The previous prime is 11000111. The next prime is 11000149. The reversal of 11000113 is 31100011.

It is a happy number.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 10131489 + 868624 = 3183^2 + 932^2 .

It is an emirp because it is prime and its reverse (31100011) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 11000113 - 2¹ = 11000111 is a prime.

Together with 11000111, it forms a pair of twin primes.

It is a junction number, because it is equal to n+sod(n) for n = 11000096 and 11000105.

It is not a weakly prime, because it can be changed into another prime (11000111) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5500056 + 5500057.

It is an arithmetic number, because the mean of its divisors is an integer number (5500057).

Almost surely, 2^11000113 is an apocalyptic number.

It is an amenable number.

11000113 is a deficient number, since it is larger than the sum of its proper divisors (1).

11000113 is an equidigital number, since it uses as much as digits as its factorization.

11000113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 3, while the sum is 7.

The square root of 11000113 is about 3316.6418257026. The cubic root of 11000113 is about 222.3987705990.

Adding to 11000113 its reverse (31100011), we get a palindrome (42100124).

It can be divided in two parts, 1100011 and 3, that multiplied together give a palindrome (3300033).

The spelling of 11000113 in words is "eleven million, one hundred thirteen", and thus it is an aban number.