21114013120033 has 4 divisors (see below), whose sum is σ = 21605036681008. Its totient is φ = 20622989559060.

The previous prime is 21114013119973. The next prime is 21114013120121. The reversal of 21114013120033 is 33002131041112.

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-21114013120033 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 21114013119986 and 21114013120013.

It is not an unprimeable number, because it can be changed into a prime (21114013127033) 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, 245511780423 + ... + 245511780508.

It is an arithmetic number, because the mean of its divisors is an integer number (5401259170252).

Almost surely, 2^{21114013120033} is an apocalyptic number.

It is an amenable number.

21114013120033 is a deficient number, since it is larger than the sum of its proper divisors (491023560975).

21114013120033 is an equidigital number, since it uses as much as digits as its factorization.

21114013120033 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 491023560974.

The product of its (nonzero) digits is 432, while the sum is 22.

Adding to 21114013120033 its reverse (33002131041112), we get a palindrome (54116144161145).

The spelling of 21114013120033 in words is "twenty-one trillion, one hundred fourteen billion, thirteen million, one hundred twenty thousand, thirty-three".