101000000113 has 2 divisors, whose sum is σ = 101000000114. Its totient is φ = 101000000112.

The previous prime is 101000000107. The next prime is 101000000137. The reversal of 101000000113 is 311000000101.

It is a happy number.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 70526362624 + 30473637489 = 265568^2 + 174567^2 .

It is a cyclic number.

It is not a de Polignac number, because 101000000113 - 2¹⁷ = 100999869041 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 101000000096 and 101000000105.

It is not a weakly prime, because it can be changed into another prime (101000000813) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50500000056 + 50500000057.

It is an arithmetic number, because the mean of its divisors is an integer number (50500000057).

Almost surely, 2^101000000113 is an apocalyptic number.

It is an amenable number.

101000000113 is a deficient number, since it is larger than the sum of its proper divisors (1).

101000000113 is an equidigital number, since it uses as much as digits as its factorization.

101000000113 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 3, while the sum is 7.

Adding to 101000000113 its reverse (311000000101), we get a palindrome (412000000214).

The spelling of 101000000113 in words is "one hundred one billion, one hundred thirteen", and thus it is an aban number.