10011520431131 has 4 divisors (see below), whose sum is σ = 10011591515064. Its totient is φ = 10011449347200.

The previous prime is 10011520431113. The next prime is 10011520431139. The reversal of 10011520431131 is 13113402511001.

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 10011520431131 - 2⁴² = 5613473920027 is a prime.

It is a super-3 number, since 3×10011520431131³ (a number of 40 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 10011520431097 and 10011520431106.

It is not an unprimeable number, because it can be changed into a prime (10011520431139) 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, 35330285 + ... + 35612526.

It is an arithmetic number, because the mean of its divisors is an integer number (2502897878766).

Almost surely, 2^{10011520431131} is an apocalyptic number.

10011520431131 is a deficient number, since it is larger than the sum of its proper divisors (71083933).

10011520431131 is an equidigital number, since it uses as much as digits as its factorization.

10011520431131 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 71083932.

The product of its (nonzero) digits is 360, while the sum is 23.

Adding to 10011520431131 its reverse (13113402511001), we get a palindrome (23124922942132).

The spelling of 10011520431131 in words is "ten trillion, eleven billion, five hundred twenty million, four hundred thirty-one thousand, one hundred thirty-one".