13000433130411 has 4 divisors (see below), whose sum is σ = 17333910840552. Its totient is φ = 8666955420272.

The previous prime is 13000433130409. The next prime is 13000433130439. The reversal of 13000433130411 is 11403133400031.

It is a semiprime because it is the product of two primes.

It is a cyclic number.

It is not a de Polignac number, because 13000433130411 - 2¹ = 13000433130409 is a prime.

It is a super-2 number, since 2×13000433130411² (a number of 27 digits) contains 22 as substring.

It is not an unprimeable number, because it can be changed into a prime (13000433134411) 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, 2166738855066 + ... + 2166738855071.

It is an arithmetic number, because the mean of its divisors is an integer number (4333477710138).

Almost surely, 2^{13000433130411} is an apocalyptic number.

13000433130411 is a deficient number, since it is larger than the sum of its proper divisors (4333477710141).

13000433130411 is an equidigital number, since it uses as much as digits as its factorization.

13000433130411 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 4333477710140.

The product of its (nonzero) digits is 1296, while the sum is 24.

Adding to 13000433130411 its reverse (11403133400031), we get a palindrome (24403566530442).

The spelling of 13000433130411 in words is "thirteen trillion, four hundred thirty-three million, one hundred thirty thousand, four hundred eleven".