115204433 has 2 divisors, whose sum is σ = 115204434. Its totient is φ = 115204432.

The previous prime is 115204409. The next prime is 115204471. The reversal of 115204433 is 334402511.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 114661264 + 543169 = 10708^2 + 737^2 .

It is an emirp because it is prime and its reverse (334402511) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 115204433 - 2²² = 111010129 is a prime.

It is a super-2 number, since 2×115204433² = 26544122765702978, which contains 22 as substring.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 115204399 and 115204408.

It is not a weakly prime, because it can be changed into another prime (115201433) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57602216 + 57602217.

It is an arithmetic number, because the mean of its divisors is an integer number (57602217).

Almost surely, 2^115204433 is an apocalyptic number.

It is an amenable number.

115204433 is a deficient number, since it is larger than the sum of its proper divisors (1).

115204433 is an equidigital number, since it uses as much as digits as its factorization.

115204433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1440, while the sum is 23.

The square root of 115204433 is about 10733.3328002070. The cubic root of 115204433 is about 486.5824008624.

Adding to 115204433 its reverse (334402511), we get a palindrome (449606944).

The spelling of 115204433 in words is "one hundred fifteen million, two hundred four thousand, four hundred thirty-three".