11324103112433 has 4 divisors (see below), whose sum is σ = 11324223612900. Its totient is φ = 11323982611968.

The previous prime is 11324103112303. The next prime is 11324103112453. The reversal of 11324103112433 is 33421130142311.

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 10296288341089 + 1027814771344 = 3208783^2 + 1013812^2 .

It is a cyclic number.

It is not a de Polignac number, because 11324103112433 - 2¹² = 11324103108337 is a prime.

It is a super-3 number, since 3×11324103112433³ (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 = 11324103112396 and 11324103112405.

It is not an unprimeable number, because it can be changed into a prime (11324103112453) 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, 60109160 + ... + 60297257.

It is an arithmetic number, because the mean of its divisors is an integer number (2831055903225).

Almost surely, 2^{11324103112433} is an apocalyptic number.

It is an amenable number.

11324103112433 is a deficient number, since it is larger than the sum of its proper divisors (120500467).

11324103112433 is an equidigital number, since it uses as much as digits as its factorization.

11324103112433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 120500466.

The product of its (nonzero) digits is 5184, while the sum is 29.

Adding to 11324103112433 its reverse (33421130142311), we get a palindrome (44745233254744).

The spelling of 11324103112433 in words is "eleven trillion, three hundred twenty-four billion, one hundred three million, one hundred twelve thousand, four hundred thirty-three".