2002143131113 has 4 divisors (see below), whose sum is σ = 2002640314960. Its totient is φ = 2001645947268.

The previous prime is 2002143131099. The next prime is 2002143131141. The reversal of 2002143131113 is 3111313412002.

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 not a de Polignac number, because 2002143131113 - 2¹³ = 2002143122921 is a prime.

It is a super-4 number, since 4×2002143131113⁴ (a number of 50 digits) contains 4444 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (2002143141113) 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, 248585883 + ... + 248593936.

It is an arithmetic number, because the mean of its divisors is an integer number (500660078740).

Almost surely, 2^{2002143131113} is an apocalyptic number.

It is an amenable number.

2002143131113 is a deficient number, since it is larger than the sum of its proper divisors (497183847).

2002143131113 is an equidigital number, since it uses as much as digits as its factorization.

2002143131113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 497183846.

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

Adding to 2002143131113 its reverse (3111313412002), we get a palindrome (5113456543115).

The spelling of 2002143131113 in words is "two trillion, two billion, one hundred forty-three million, one hundred thirty-one thousand, one hundred thirteen".