125041200433 has 4 divisors (see below), whose sum is σ = 125042493940. Its totient is φ = 125039906928.

The previous prime is 125041200407. The next prime is 125041200437. The reversal of 125041200433 is 334002140521.

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

It can be written as a sum of positive squares in 2 ways, for example, as 103937246449 + 21103953984 = 322393^2 + 145272^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-125041200433 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 125041200398 and 125041200407.

It is not an unprimeable number, because it can be changed into a prime (125041200437) 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, 488910 + ... + 699367.

It is an arithmetic number, because the mean of its divisors is an integer number (31260623485).

Almost surely, 2^125041200433 is an apocalyptic number.

It is an amenable number.

125041200433 is a deficient number, since it is larger than the sum of its proper divisors (1293507).

125041200433 is a wasteful number, since it uses less digits than its factorization.

125041200433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1293506.

The product of its (nonzero) digits is 2880, while the sum is 25.

Adding to 125041200433 its reverse (334002140521), we get a palindrome (459043340954).

The spelling of 125041200433 in words is "one hundred twenty-five billion, forty-one million, two hundred thousand, four hundred thirty-three".