10040100113 has 2 divisors, whose sum is σ = 10040100114. Its totient is φ = 10040100112.

The previous prime is 10040100089. The next prime is 10040100127. The reversal of 10040100113 is 31100104001.

10040100113 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 5762024464 + 4278075649 = 75908^2 + 65407^2 .

It is a cyclic number.

It is not a de Polignac number, because 10040100113 - 2¹⁶ = 10040034577 is a prime.

It is a super-2 number, since 2×10040100113² (a number of 21 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 10040100094 and 10040100103.

It is not a weakly prime, because it can be changed into another prime (10040700113) 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 as a sum of consecutive naturals, namely, 5020050056 + 5020050057.

It is an arithmetic number, because the mean of its divisors is an integer number (5020050057).

Almost surely, 2^10040100113 is an apocalyptic number.

It is an amenable number.

10040100113 is a deficient number, since it is larger than the sum of its proper divisors (1).

10040100113 is an equidigital number, since it uses as much as digits as its factorization.

10040100113 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 12, while the sum is 11.

Adding to 10040100113 its reverse (31100104001), we get a palindrome (41140204114).

The spelling of 10040100113 in words is "ten billion, forty million, one hundred thousand, one hundred thirteen".