11000433130301 has 2 divisors, whose sum is σ = 11000433130302. Its totient is φ = 11000433130300.

The previous prime is 11000433130229. The next prime is 11000433130363. The reversal of 11000433130301 is 10303133400011.

11000433130301 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 8080090502500 + 2920342627801 = 2842550^2 + 1708901^2 .

It is an emirp because it is prime and its reverse (10303133400011) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-11000433130301 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (11000433130201) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5500216565150 + 5500216565151.

It is an arithmetic number, because the mean of its divisors is an integer number (5500216565151).

Almost surely, 2^{11000433130301} is an apocalyptic number.

It is an amenable number.

11000433130301 is a deficient number, since it is larger than the sum of its proper divisors (1).

11000433130301 is an equidigital number, since it uses as much as digits as its factorization.

11000433130301 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 324, while the sum is 20.

Adding to 11000433130301 its reverse (10303133400011), we get a palindrome (21303566530312).

The spelling of 11000433130301 in words is "eleven trillion, four hundred thirty-three million, one hundred thirty thousand, three hundred one".