12143443153 has 4 divisors (see below), whose sum is σ = 12782571760. Its totient is φ = 11504314548.

The previous prime is 12143443139. The next prime is 12143443177. The reversal of 12143443153 is 35134434121.

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, and also an emirpimes, since its reverse is a distinct semiprime: 35134434121 = 1361 ⋅25815161.

It is a cyclic number.

It is not a de Polignac number, because 12143443153 - 2¹⁷ = 12143312081 is a prime.

It is a super-3 number, since 3×12143443153³ (a number of 31 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (12143443133) 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 3 ways as a sum of consecutive naturals, for example, 319564275 + ... + 319564312.

It is an arithmetic number, because the mean of its divisors is an integer number (3195642940).

Almost surely, 2^12143443153 is an apocalyptic number.

It is an amenable number.

12143443153 is a deficient number, since it is larger than the sum of its proper divisors (639128607).

12143443153 is an equidigital number, since it uses as much as digits as its factorization.

12143443153 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 639128606.

The product of its digits is 17280, while the sum is 31.

Adding to 12143443153 its reverse (35134434121), we get a palindrome (47277877274).

The spelling of 12143443153 in words is "twelve billion, one hundred forty-three million, four hundred forty-three thousand, one hundred fifty-three".