1101111020113 has 4 divisors (see below), whose sum is σ = 1185811867828. Its totient is φ = 1016410172400.

The previous prime is 1101111020099. The next prime is 1101111020129. The reversal of 1101111020113 is 3110201111011.

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 267710003649 + 833401016464 = 517407^2 + 912908^2 .

It is a cyclic number.

It is not a de Polignac number, because 1101111020113 - 2³³ = 1092521085521 is a prime.

It is a super-3 number, since 3×1101111020113³ (a number of 37 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a Harshad number since it is a multiple of its sum of digits (13), and also a Moran number because the ratio is a prime number: 84700847701 = 1101111020113 / (1 + 1 + 0 + 1 + 1 + 1 + 1 + 0 + 2 + 0 + 1 + 1 + 3).

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 1101111020093 and 1101111020102.

It is not an unprimeable number, because it can be changed into a prime (1101111021113) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 42350423838 + ... + 42350423863.

It is an arithmetic number, because the mean of its divisors is an integer number (296452966957).

Almost surely, 2^{1101111020113} is an apocalyptic number.

1101111020113 is a gapful number since it is divisible by the number (13) formed by its first and last digit.

It is an amenable number.

1101111020113 is a deficient number, since it is larger than the sum of its proper divisors (84700847715).

1101111020113 is an equidigital number, since it uses as much as digits as its factorization.

1101111020113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 84700847714.

The product of its (nonzero) digits is 6, while the sum is 13.

Adding to 1101111020113 its reverse (3110201111011), we get a palindrome (4211312131124).

The spelling of 1101111020113 in words is "one trillion, one hundred one billion, one hundred eleven million, twenty thousand, one hundred thirteen".