115134433 has 4 divisors (see below), whose sum is σ = 121194160. Its totient is φ = 109074708.

The previous prime is 115134431. The next prime is 115134451. The reversal of 115134433 is 334431511.

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.

It is a cyclic number.

It is not a de Polignac number, because 115134433 - 2¹ = 115134431 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 115134398 and 115134407.

It is not an unprimeable number, because it can be changed into a prime (115134431) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 3029835 + ... + 3029872.

It is an arithmetic number, because the mean of its divisors is an integer number (30298540).

Almost surely, 2^115134433 is an apocalyptic number.

It is an amenable number.

115134433 is a deficient number, since it is larger than the sum of its proper divisors (6059727).

115134433 is an equidigital number, since it uses as much as digits as its factorization.

115134433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 6059726.

The product of its digits is 2160, while the sum is 25.

The square root of 115134433 is about 10730.0714349905. The cubic root of 115134433 is about 486.4838292244.

Adding to 115134433 its reverse (334431511), we get a palindrome (449565944).

The spelling of 115134433 in words is "one hundred fifteen million, one hundred thirty-four thousand, four hundred thirty-three".