125403433 has 4 divisors (see below), whose sum is σ = 125653248. Its totient is φ = 125153620.

The previous prime is 125403391. The next prime is 125403451. The reversal of 125403433 is 334304521.

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 125403433 - 2¹⁷ = 125272361 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 125403398 and 125403407.

It is not an unprimeable number, because it can be changed into a prime (125403493) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 124153 + ... + 125158.

It is an arithmetic number, because the mean of its divisors is an integer number (31413312).

Almost surely, 2^125403433 is an apocalyptic number.

It is an amenable number.

125403433 is a deficient number, since it is larger than the sum of its proper divisors (249815).

125403433 is an equidigital number, since it uses as much as digits as its factorization.

125403433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 249814.

The product of its (nonzero) digits is 4320, while the sum is 25.

The square root of 125403433 is about 11198.3674256563. The cubic root of 125403433 is about 500.5373330063.

Adding to 125403433 its reverse (334304521), we get a palindrome (459707954).

The spelling of 125403433 in words is "one hundred twenty-five million, four hundred three thousand, four hundred thirty-three".