13001152555535 has 4 divisors (see below), whose sum is σ = 15601383066648. Its totient is φ = 10400922044424.

The previous prime is 13001152555513. The next prime is 13001152555571. The reversal of 13001152555535 is 53555525110031.

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

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-13001152555535 is a prime.

It is a super-2 number, since 2×13001152555535² (a number of 27 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 = 13001152555492 and 13001152555501.

It is a congruent number.

It is an unprimeable number.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1300115255549 + ... + 1300115255558.

It is an arithmetic number, because the mean of its divisors is an integer number (3900345766662).

Almost surely, 2^{13001152555535} is an apocalyptic number.

13001152555535 is a deficient number, since it is larger than the sum of its proper divisors (2600230511113).

13001152555535 is an equidigital number, since it uses as much as digits as its factorization.

13001152555535 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2600230511112.

The product of its (nonzero) digits is 281250, while the sum is 41.

Adding to 13001152555535 its reverse (53555525110031), we get a palindrome (66556677665566).

The spelling of 13001152555535 in words is "thirteen trillion, one billion, one hundred fifty-two million, five hundred fifty-five thousand, five hundred thirty-five".