1100031113 has 2 divisors, whose sum is σ = 1100031114. Its totient is φ = 1100031112.

The previous prime is 1100031089. The next prime is 1100031131. The reversal of 1100031113 is 3111300011.

It is a happy number.

Together with next prime (1100031131) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 891141904 + 208889209 = 29852^2 + 14453^2 .

It is an emirp because it is prime and its reverse (3111300011) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1100031113 - 2¹⁰ = 1100030089 is a prime.

It is a super-3 number, since 3×1100031113³ (a number of 28 digits) contains 333 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 1100031094 and 1100031103.

It is not a weakly prime, because it can be changed into another prime (1100031133) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 550015556 + 550015557.

It is an arithmetic number, because the mean of its divisors is an integer number (550015557).

Almost surely, 2^1100031113 is an apocalyptic number.

It is an amenable number.

1100031113 is a deficient number, since it is larger than the sum of its proper divisors (1).

1100031113 is an equidigital number, since it uses as much as digits as its factorization.

1100031113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 9, while the sum is 11.

The square root of 1100031113 is about 33166.7169463605. The cubic root of 1100031113 is about 1032.2898478892.

Adding to 1100031113 its reverse (3111300011), we get a palindrome (4211331124).

The spelling of 1100031113 in words is "one billion, one hundred million, thirty-one thousand, one hundred thirteen".