100100000000111 has 2 divisors, whose sum is σ = 100100000000112. Its totient is φ = 100100000000110.

The previous prime is 100100000000093. The next prime is 100100000000113. The reversal of 100100000000111 is 111000000001001.

It is a strong prime.

It is an emirp because it is prime and its reverse (111000000001001) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 100100000000111 - 2³⁴ = 100082820130927 is a prime.

Together with 100100000000113, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 100100000000095 and 100100000000104.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (100100000000113) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50050000000055 + 50050000000056.

It is an arithmetic number, because the mean of its divisors is an integer number (50050000000056).

Almost surely, 2^{100100000000111} is an apocalyptic number.

100100000000111 is a deficient number, since it is larger than the sum of its proper divisors (1).

100100000000111 is an equidigital number, since it uses as much as digits as its factorization.

100100000000111 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1, while the sum is 5.

Adding to 100100000000111 its reverse (111000000001001), we get a palindrome (211100000001112).

The spelling of 100100000000111 in words is "one hundred trillion, one hundred billion, one hundred eleven", and thus it is an aban number.