1100433130121 has 4 divisors (see below), whose sum is σ = 1185081832452. Its totient is φ = 1015784427792.

The previous prime is 1100433130109. The next prime is 1100433130129. The reversal of 1100433130121 is 1210313340011.

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 249960001600 + 850473128521 = 499960^2 + 922211^2 .

It is a cyclic number.

It is not a de Polignac number, because 1100433130121 - 2¹⁴ = 1100433113737 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 1100433130093 and 1100433130102.

It is not an unprimeable number, because it can be changed into a prime (1100433130129) 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, 42324351146 + ... + 42324351171.

It is an arithmetic number, because the mean of its divisors is an integer number (296270458113).

Almost surely, 2^{1100433130121} is an apocalyptic number.

It is an amenable number.

1100433130121 is a deficient number, since it is larger than the sum of its proper divisors (84648702331).

1100433130121 is an equidigital number, since it uses as much as digits as its factorization.

1100433130121 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 84648702330.

The product of its (nonzero) digits is 216, while the sum is 20.

Adding to 1100433130121 its reverse (1210313340011), we get a palindrome (2310746470132).

The spelling of 1100433130121 in words is "one trillion, one hundred billion, four hundred thirty-three million, one hundred thirty thousand, one hundred twenty-one".