11002324433 has 6 divisors (see below), whose sum is σ = 11611871874. Its totient is φ = 10423254384.

The previous prime is 11002324427. The next prime is 11002324453. The reversal of 11002324433 is 33442320011.

11002324433 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It can be written as a sum of positive squares in only one way, i.e., 6211331344 + 4790993089 = 78812^2 + 69217^2 .

It is not a de Polignac number, because 11002324433 - 2¹⁸ = 11002062289 is a prime.

It is a super-4 number, since 4×11002324433⁴ (a number of 41 digits) contains 4444 as substring. Note that it is a super-d number also for d = 2.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 11002324399 and 11002324408.

It is not an unprimeable number, because it can be changed into a prime (11002324453) 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 5 ways as a sum of consecutive naturals, for example, 15238316 + ... + 15239037.

It is an arithmetic number, because the mean of its divisors is an integer number (1935311979).

Almost surely, 2^11002324433 is an apocalyptic number.

It is an amenable number.

11002324433 is a deficient number, since it is larger than the sum of its proper divisors (609547441).

11002324433 is an equidigital number, since it uses as much as digits as its factorization.

11002324433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 30477391 (or 30477372 counting only the distinct ones).

The product of its (nonzero) digits is 1728, while the sum is 23.

Adding to 11002324433 its reverse (33442320011), we get a palindrome (44444644444).

The spelling of 11002324433 in words is "eleven billion, two million, three hundred twenty-four thousand, four hundred thirty-three".