131113101130230 has 64 divisors (see below), whose sum is σ = 331313675904000. Its totient is φ = 33115251928320.

The previous prime is 131113101130201. The next prime is 131113101130237. The reversal of 131113101130230 is 32031101311131.

It is a Curzon number.

It is a junction number, because it is equal to n+sod(n) for n = 131113101130197 and 131113101130206.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (131113101130237) by changing a digit.

It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 25633255 + ... + 30319794.

It is an arithmetic number, because the mean of its divisors is an integer number (5176776186000).

Almost surely, 2^{131113101130230} is an apocalyptic number.

131113101130230 is a gapful number since it is divisible by the number (10) formed by its first and last digit.

131113101130230 is an abundant number, since it is smaller than the sum of its proper divisors (200200574773770).

It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.

131113101130230 is a wasteful number, since it uses less digits than its factorization.

131113101130230 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 55957189.

The product of its (nonzero) digits is 162, while the sum is 21.

Adding to 131113101130230 its reverse (32031101311131), we get a palindrome (163144202441361).

The spelling of 131113101130230 in words is "one hundred thirty-one trillion, one hundred thirteen billion, one hundred one million, one hundred thirty thousand, two hundred thirty".