11001100103113 has 2 divisors, whose sum is σ = 11001100103114. Its totient is φ = 11001100103112.

The previous prime is 11001100103053. The next prime is 11001100103131. The reversal of 11001100103113 is 31130100110011.

Together with next prime (11001100103131) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 5550985738809 + 5450114364304 = 2356053^2 + 2334548^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-11001100103113 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 11001100103093 and 11001100103102.

It is not a weakly prime, because it can be changed into another prime (11001103103113) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5500550051556 + 5500550051557.

It is an arithmetic number, because the mean of its divisors is an integer number (5500550051557).

Almost surely, 2^{11001100103113} is an apocalyptic number.

It is an amenable number.

11001100103113 is a deficient number, since it is larger than the sum of its proper divisors (1).

11001100103113 is an equidigital number, since it uses as much as digits as its factorization.

11001100103113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 9, while the sum is 13.

Adding to 11001100103113 its reverse (31130100110011), we get a palindrome (42131200213124).

The spelling of 11001100103113 in words is "eleven trillion, one billion, one hundred million, one hundred three thousand, one hundred thirteen".