12043315015433 has 4 divisors (see below), whose sum is σ = 12969723862788. Its totient is φ = 11116906168080.

The previous prime is 12043315015411. The next prime is 12043315015441. The reversal of 12043315015433 is 33451051334021.

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 3436678253584 + 8606636761849 = 1853828^2 + 2933707^2 .

It is a cyclic number.

It is not a de Polignac number, because 12043315015433 - 2⁸ = 12043315015177 is a prime.

It is a super-2 number, since 2×12043315015433² (a number of 27 digits) contains 22 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 12043315015393 and 12043315015402.

It is not an unprimeable number, because it can be changed into a prime (12043315015483) 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, 463204423658 + ... + 463204423683.

It is an arithmetic number, because the mean of its divisors is an integer number (3242430965697).

Almost surely, 2^{12043315015433} is an apocalyptic number.

12043315015433 is a gapful number since it is divisible by the number (13) formed by its first and last digit.

It is an amenable number.

12043315015433 is a deficient number, since it is larger than the sum of its proper divisors (926408847355).

12043315015433 is an equidigital number, since it uses as much as digits as its factorization.

12043315015433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 926408847354.

The product of its (nonzero) digits is 64800, while the sum is 35.

Adding to 12043315015433 its reverse (33451051334021), we get a palindrome (45494366349454).

The spelling of 12043315015433 in words is "twelve trillion, forty-three billion, three hundred fifteen million, fifteen thousand, four hundred thirty-three".