100130035 has 4 divisors (see below), whose sum is σ = 120156048. Its totient is φ = 80104024.

The previous prime is 100130003. The next prime is 100130039. The reversal of 100130035 is 530031001.

It is a semiprime because it is the product of two primes.

It is a cyclic number.

It is not a de Polignac number, because 100130035 - 2⁵ = 100130003 is a prime.

It is a super-3 number, since 3×100130035³ (a number of 25 digits) contains 333 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 100129997 and 100130024.

It is not an unprimeable number, because it can be changed into a prime (100130039) 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, 10012999 + ... + 10013008.

It is an arithmetic number, because the mean of its divisors is an integer number (30039012).

Almost surely, 2^100130035 is an apocalyptic number.

100130035 is a deficient number, since it is larger than the sum of its proper divisors (20026013).

100130035 is an equidigital number, since it uses as much as digits as its factorization.

100130035 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 20026012.

The product of its (nonzero) digits is 45, while the sum is 13.

The square root of 100130035 is about 10006.4996377355. The cubic root of 100130035 is about 464.3599858866.

Adding to 100130035 its reverse (530031001), we get a palindrome (630161036).

The spelling of 100130035 in words is "one hundred million, one hundred thirty thousand, thirty-five".