101131131113 has 4 divisors (see below), whose sum is σ = 101927439360. Its totient is φ = 100334822868.

The previous prime is 101131131103. The next prime is 101131131119. The reversal of 101131131113 is 311131131101.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.

It is a cyclic number.

It is not a de Polignac number, because 101131131113 - 2⁸ = 101131130857 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 101131131091 and 101131131100.

It is not an unprimeable number, because it can be changed into a prime (101131131119) 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, 398153933 + ... + 398154186.

It is an arithmetic number, because the mean of its divisors is an integer number (25481859840).

Almost surely, 2^101131131113 is an apocalyptic number.

It is an amenable number.

101131131113 is a deficient number, since it is larger than the sum of its proper divisors (796308247).

101131131113 is an equidigital number, since it uses as much as digits as its factorization.

101131131113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 796308246.

The product of its (nonzero) digits is 27, while the sum is 17.

Adding to 101131131113 its reverse (311131131101), we get a palindrome (412262262214).

The spelling of 101131131113 in words is "one hundred one billion, one hundred thirty-one million, one hundred thirty-one thousand, one hundred thirteen".