11004031143133 has 8 divisors (see below), whose sum is σ = 12576092431680. Its totient is φ = 9431984064480.

The previous prime is 11004031143037. The next prime is 11004031143137. The reversal of 11004031143133 is 33134113040011.

It is a sphenic number, since it is the product of 3 distinct primes.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-11004031143133 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 11004031143098 and 11004031143107.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (11004031143137) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 1837888 + ... + 5038441.

It is an arithmetic number, because the mean of its divisors is an integer number (1572011553960).

Almost surely, 2^{11004031143133} is an apocalyptic number.

It is an amenable number.

11004031143133 is a deficient number, since it is larger than the sum of its proper divisors (1572061288547).

11004031143133 is an equidigital number, since it uses as much as digits as its factorization.

11004031143133 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 7104947.

The product of its (nonzero) digits is 1296, while the sum is 25.

Adding to 11004031143133 its reverse (33134113040011), we get a palindrome (44138144183144).

The spelling of 11004031143133 in words is "eleven trillion, four billion, thirty-one million, one hundred forty-three thousand, one hundred thirty-three".