115111103 has 2 divisors, whose sum is σ = 115111104. Its totient is φ = 115111102.

The previous prime is 115111069. The next prime is 115111111. The reversal of 115111103 is 301111511.

It is a strong prime.

It is an emirp because it is prime and its reverse (301111511) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 115111103 - 2¹² = 115107007 is a prime.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (115111163) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57555551 + 57555552.

It is an arithmetic number, because the mean of its divisors is an integer number (57555552).

Almost surely, 2^115111103 is an apocalyptic number.

115111103 is a deficient number, since it is larger than the sum of its proper divisors (1).

115111103 is an equidigital number, since it uses as much as digits as its factorization.

115111103 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 15, while the sum is 14.

The square root of 115111103 is about 10728.9842482874. The cubic root of 115111103 is about 486.4509678289.

Adding to 115111103 its reverse (301111511), we get a palindrome (416222614).

The spelling of 115111103 in words is "one hundred fifteen million, one hundred eleven thousand, one hundred three".