1000100113 has 4 divisors (see below), whose sum is σ = 1004250148. Its totient is φ = 995950080.

The previous prime is 1000100077. The next prime is 1000100161. The reversal of 1000100113 is 3110010001.

It is a happy number.

It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 3110010001 = 7 ⋅444287143.

It can be written as a sum of positive squares in 2 ways, for example, as 712729809 + 287370304 = 26697^2 + 16952^2 .

It is a cyclic number.

It is not a de Polignac number, because 1000100113 - 2⁹ = 1000099601 is a prime.

It is a super-2 number, since 2×1000100113² = 2000400472045225538, which contains 22 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 1000100096 and 1000100105.

It is not an unprimeable number, because it can be changed into a prime (1000100173) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2074656 + ... + 2075137.

It is an arithmetic number, because the mean of its divisors is an integer number (251062537).

Almost surely, 2^1000100113 is an apocalyptic number.

It is an amenable number.

1000100113 is a deficient number, since it is larger than the sum of its proper divisors (4150035).

1000100113 is an equidigital number, since it uses as much as digits as its factorization.

1000100113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 4150034.

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

The square root of 1000100113 is about 31624.3594875849. The cubic root of 1000100113 is about 1000.0333698864.

Adding to 1000100113 its reverse (3110010001), we get a palindrome (4110110114).

The spelling of 1000100113 in words is "one billion, one hundred thousand, one hundred thirteen".