13544053433 has 4 divisors (see below), whose sum is σ = 13544389860. Its totient is φ = 13543717008.

The previous prime is 13544053409. The next prime is 13544053469. The reversal of 13544053433 is 33435044531.

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 13148520889 + 395532544 = 114667^2 + 19888^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-13544053433 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 13544053393 and 13544053402.

It is not an unprimeable number, because it can be changed into a prime (13544056433) 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, 98078 + ... + 191591.

It is an arithmetic number, because the mean of its divisors is an integer number (3386097465).

Almost surely, 2^13544053433 is an apocalyptic number.

It is an amenable number.

13544053433 is a deficient number, since it is larger than the sum of its proper divisors (336427).

13544053433 is an equidigital number, since it uses as much as digits as its factorization.

13544053433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 336426.

The product of its (nonzero) digits is 129600, while the sum is 35.

Adding to 13544053433 its reverse (33435044531), we get a palindrome (46979097964).

The spelling of 13544053433 in words is "thirteen billion, five hundred forty-four million, fifty-three thousand, four hundred thirty-three".