11011307 has 2 divisors, whose sum is σ = 11011308. Its totient is φ = 11011306.

The previous prime is 11011303. The next prime is 11011321. The reversal of 11011307 is 70311011.

It is an a-pointer prime, because the next prime (11011321) can be obtained adding 11011307 to its sum of digits (14).

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 11011307 - 2² = 11011303 is a prime.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 11011291 and 11011300.

It is not a weakly prime, because it can be changed into another prime (11011303) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5505653 + 5505654.

It is an arithmetic number, because the mean of its divisors is an integer number (5505654).

Almost surely, 2^11011307 is an apocalyptic number.

11011307 is a deficient number, since it is larger than the sum of its proper divisors (1).

11011307 is an equidigital number, since it uses as much as digits as its factorization.

11011307 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 21, while the sum is 14.

The square root of 11011307 is about 3318.3289469249. The cubic root of 11011307 is about 222.4741846075.

Adding to 11011307 its reverse (70311011), we get a palindrome (81322318).

The spelling of 11011307 in words is "eleven million, eleven thousand, three hundred seven".