5304000115 has 4 divisors (see below), whose sum is σ = 6364800144. Its totient is φ = 4243200088.

The previous prime is 5304000103. The next prime is 5304000131. The reversal of 5304000115 is 5110004035.

5304000115 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.

It is a semiprime because it is the product of two primes.

It is a cyclic number.

It is not a de Polignac number, because 5304000115 - 2⁵ = 5304000083 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 5304000092 and 5304000101.

It is an unprimeable number.

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, 530400007 + ... + 530400016.

It is an arithmetic number, because the mean of its divisors is an integer number (1591200036).

Almost surely, 2^5304000115 is an apocalyptic number.

5304000115 is a deficient number, since it is larger than the sum of its proper divisors (1060800029).

5304000115 is a wasteful number, since it uses less digits than its factorization.

5304000115 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1060800028.

The product of its (nonzero) digits is 300, while the sum is 19.

The square root of 5304000115 is about 72828.5666136579. The cubic root of 5304000115 is about 1743.9519232944.

The spelling of 5304000115 in words is "five billion, three hundred four million, one hundred fifteen", and thus it is an aban number.