110004332324441 has 4 divisors (see below), whose sum is σ = 110004743898384. Its totient is φ = 110003920750500.

The previous prime is 110004332324423. The next prime is 110004332324443. The reversal of 110004332324441 is 144423233400011.

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 a de Polignac number, because none of the positive numbers 2^k-110004332324441 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 110004332324398 and 110004332324407.

It is not an unprimeable number, because it can be changed into a prime (110004332324443) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 205385795 + ... + 205920696.

It is an arithmetic number, because the mean of its divisors is an integer number (27501185974596).

Almost surely, 2^{110004332324441} is an apocalyptic number.

It is an amenable number.

110004332324441 is a deficient number, since it is larger than the sum of its proper divisors (411573943).

110004332324441 is an equidigital number, since it uses as much as digits as its factorization.

110004332324441 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 411573942.

The product of its (nonzero) digits is 27648, while the sum is 32.

Adding to 110004332324441 its reverse (144423233400011), we get a palindrome (254427565724452).

The spelling of 110004332324441 in words is "one hundred ten trillion, four billion, three hundred thirty-two million, three hundred twenty-four thousand, four hundred forty-one".