100121003113 has 2 divisors, whose sum is σ = 100121003114. Its totient is φ = 100121003112.

The previous prime is 100121003089. The next prime is 100121003131. The reversal of 100121003113 is 311300121001.

Together with next prime (100121003131) 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., 71039840089 + 29081163024 = 266533^2 + 170532^2 .

It is a cyclic number.

It is not a de Polignac number, because 100121003113 - 2⁹ = 100121002601 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 100121003093 and 100121003102.

It is not a weakly prime, because it can be changed into another prime (100121033113) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50060501556 + 50060501557.

It is an arithmetic number, because the mean of its divisors is an integer number (50060501557).

Almost surely, 2^100121003113 is an apocalyptic number.

It is an amenable number.

100121003113 is a deficient number, since it is larger than the sum of its proper divisors (1).

100121003113 is an equidigital number, since it uses as much as digits as its factorization.

100121003113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 18, while the sum is 13.

Adding to 100121003113 its reverse (311300121001), we get a palindrome (411421124114).

The spelling of 100121003113 in words is "one hundred billion, one hundred twenty-one million, three thousand, one hundred thirteen".