131012224140433 has 4 divisors (see below), whose sum is σ = 131012263457460. Its totient is φ = 131012184823408.

The previous prime is 131012224140383. The next prime is 131012224140491. The reversal of 131012224140433 is 334041422210131.

It is a happy number.

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

It can be written as a sum of positive squares in 2 ways, for example, as 51319056167824 + 79693167972609 = 7163732^2 + 8927103^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-131012224140433 is a prime.

It is a super-4 number, since 4×131012224140433⁴ (a number of 58 digits) contains 4444 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 131012224140395 and 131012224140404.

It is not an unprimeable number, because it can be changed into a prime (131012224170433) 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, 14144710 + ... + 21496447.

It is an arithmetic number, because the mean of its divisors is an integer number (32753065864365).

Almost surely, 2^{131012224140433} is an apocalyptic number.

It is an amenable number.

131012224140433 is a deficient number, since it is larger than the sum of its proper divisors (39317027).

131012224140433 is an equidigital number, since it uses as much as digits as its factorization.

131012224140433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 39317026.

The product of its (nonzero) digits is 13824, while the sum is 31.

Adding to 131012224140433 its reverse (334041422210131), we get a palindrome (465053646350564).

The spelling of 131012224140433 in words is "one hundred thirty-one trillion, twelve billion, two hundred twenty-four million, one hundred forty thousand, four hundred thirty-three".