53143131121 has 4 divisors (see below), whose sum is σ = 53164130544. Its totient is φ = 53122131700.

The previous prime is 53143131109. The next prime is 53143131211. The reversal of 53143131121 is 12113134135.

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: 12113134135 = 5 ⋅2422626827.

It is a cyclic number.

It is not a de Polignac number, because 53143131121 - 2¹¹ = 53143129073 is a prime.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (53143131101) 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, 10495915 + ... + 10500976.

It is an arithmetic number, because the mean of its divisors is an integer number (13291032636).

Almost surely, 2^53143131121 is an apocalyptic number.

It is an amenable number.

53143131121 is a deficient number, since it is larger than the sum of its proper divisors (20999423).

53143131121 is a wasteful number, since it uses less digits than its factorization.

53143131121 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 20999422.

The product of its digits is 1080, while the sum is 25.

Adding to 53143131121 its reverse (12113134135), we get a palindrome (65256265256).

The spelling of 53143131121 in words is "fifty-three billion, one hundred forty-three million, one hundred thirty-one thousand, one hundred twenty-one".