13111514052115 has 4 divisors (see below), whose sum is σ = 15733816862544. Its totient is φ = 10489211241688.

The previous prime is 13111514052097. The next prime is 13111514052121. The reversal of 13111514052115 is 51125041511131.

13111514052115 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 is a cyclic number.

It is not a de Polignac number, because 13111514052115 - 2¹¹ = 13111514050067 is a prime.

It is a super-3 number, since 3×13111514052115³ (a number of 40 digits) contains 333 as substring.

It is a Duffinian 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, 1311151405207 + ... + 1311151405216.

It is an arithmetic number, because the mean of its divisors is an integer number (3933454215636).

Almost surely, 2^{13111514052115} is an apocalyptic number.

13111514052115 is a deficient number, since it is larger than the sum of its proper divisors (2622302810429).

13111514052115 is an equidigital number, since it uses as much as digits as its factorization.

13111514052115 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2622302810428.

The product of its (nonzero) digits is 3000, while the sum is 31.

Adding to 13111514052115 its reverse (51125041511131), we get a palindrome (64236555563246).

The spelling of 13111514052115 in words is "thirteen trillion, one hundred eleven billion, five hundred fourteen million, fifty-two thousand, one hundred fifteen".