101040113 has 4 divisors (see below), whose sum is σ = 103504548. Its totient is φ = 98575680.

The previous prime is 101040077. The next prime is 101040139. The reversal of 101040113 is 311040101.

It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 311040101 = 6823 ⋅45587.

It can be written as a sum of positive squares in 2 ways, for example, as 21687649 + 79352464 = 4657^2 + 8908^2 .

It is a cyclic number.

It is not a de Polignac number, because 101040113 - 2²² = 96845809 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 101040094 and 101040103.

It is not an unprimeable number, because it can be changed into a prime (101040143) 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, 1232156 + ... + 1232237.

It is an arithmetic number, because the mean of its divisors is an integer number (25876137).

Almost surely, 2^101040113 is an apocalyptic number.

It is an amenable number.

101040113 is a deficient number, since it is larger than the sum of its proper divisors (2464435).

101040113 is an equidigital number, since it uses as much as digits as its factorization.

101040113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 2464434.

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

The square root of 101040113 is about 10051.8711193489. The cubic root of 101040113 is about 465.7625949709.

Subtracting from 101040113 its product of nonzero digits (12), we obtain a palindrome (101040101).

Adding to 101040113 its reverse (311040101), we get a palindrome (412080214).

The spelling of 101040113 in words is "one hundred one million, forty thousand, one hundred thirteen".