11410023012433 has 4 divisors (see below), whose sum is σ = 11420756807488. Its totient is φ = 11399289217380.

The previous prime is 11410023012413. The next prime is 11410023012473. The reversal of 11410023012433 is 33421032001411.

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 11410023012433 - 2²⁵ = 11409989458001 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 11410023012398 and 11410023012407.

It is not an unprimeable number, because it can be changed into a prime (11410023012413) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 5366895933 + ... + 5366898058.

It is an arithmetic number, because the mean of its divisors is an integer number (2855189201872).

Almost surely, 2^{11410023012433} is an apocalyptic number.

It is an amenable number.

11410023012433 is a deficient number, since it is larger than the sum of its proper divisors (10733795055).

11410023012433 is a wasteful number, since it uses less digits than its factorization.

11410023012433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 10733795054.

The product of its (nonzero) digits is 1728, while the sum is 25.

Adding to 11410023012433 its reverse (33421032001411), we get a palindrome (44831055013844).

The spelling of 11410023012433 in words is "eleven trillion, four hundred ten billion, twenty-three million, twelve thousand, four hundred thirty-three".