53715131149 has 4 divisors (see below), whose sum is σ = 56874844764. Its totient is φ = 50555417536.

The previous prime is 53715131147. The next prime is 53715131207. The reversal of 53715131149 is 94113151735.

53715131149 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

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

It can be written as a sum of positive squares in 2 ways, for example, as 35370324900 + 18344806249 = 188070^2 + 135443^2 .

It is a cyclic number.

It is not a de Polignac number, because 53715131149 - 2¹ = 53715131147 is a prime.

It is a super-3 number, since 3×53715131149³ (a number of 33 digits) contains 333 as substring.

It is a Duffinian number.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (53715131143) 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, 1579856782 + ... + 1579856815.

It is an arithmetic number, because the mean of its divisors is an integer number (14218711191).

Almost surely, 2^53715131149 is an apocalyptic number.

It is an amenable number.

53715131149 is a deficient number, since it is larger than the sum of its proper divisors (3159713615).

53715131149 is a wasteful number, since it uses less digits than its factorization.

53715131149 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 3159713614.

The product of its digits is 56700, while the sum is 40.

The spelling of 53715131149 in words is "fifty-three billion, seven hundred fifteen million, one hundred thirty-one thousand, one hundred forty-nine".