144035112113533 has 4 divisors (see below), whose sum is σ = 144035139797500. Its totient is φ = 144035084429568.

The previous prime is 144035112113531. The next prime is 144035112113551. The reversal of 144035112113533 is 335311211530441.

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 99908081132409 + 44127030981124 = 9995403^2 + 6642818^2 .

It is a cyclic number.

It is not a de Polignac number, because 144035112113533 - 2¹ = 144035112113531 is a prime.

It is a super-2 number, since 2×144035112113533² (a number of 29 digits) contains 22 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 144035112113492 and 144035112113501.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (144035112113531) 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, 3424158 + ... + 17314591.

It is an arithmetic number, because the mean of its divisors is an integer number (36008784949375).

Almost surely, 2^{144035112113533} is an apocalyptic number.

It is an amenable number.

144035112113533 is a deficient number, since it is larger than the sum of its proper divisors (27683967).

144035112113533 is an equidigital number, since it uses as much as digits as its factorization.

144035112113533 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 27683966.

The product of its (nonzero) digits is 64800, while the sum is 37.

Adding to 144035112113533 its reverse (335311211530441), we get a palindrome (479346323643974).

The spelling of 144035112113533 in words is "one hundred forty-four trillion, thirty-five billion, one hundred twelve million, one hundred thirteen thousand, five hundred thirty-three".