5512114102433 has 4 divisors (see below), whose sum is σ = 5512152046740. Its totient is φ = 5512076158128.

The previous prime is 5512114102337. The next prime is 5512114102477. The reversal of 5512114102433 is 3342014112155.

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 1434573921169 + 4077540181264 = 1197737^2 + 2019292^2 .

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-5512114102433 is a prime.

It is a super-3 number, since 3×5512114102433³ (a number of 39 digits) contains 333 as substring.

It is a Duffinian number.

It is a self number, because there is not a number n which added to its sum of digits gives 5512114102433.

It is not an unprimeable number, because it can be changed into a prime (5512114102483) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 18753410 + ... + 19045067.

It is an arithmetic number, because the mean of its divisors is an integer number (1378038011685).

Almost surely, 2^{5512114102433} is an apocalyptic number.

It is an amenable number.

5512114102433 is a deficient number, since it is larger than the sum of its proper divisors (37944307).

5512114102433 is a wasteful number, since it uses less digits than its factorization.

5512114102433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 37944306.

The product of its (nonzero) digits is 14400, while the sum is 32.

Adding to 5512114102433 its reverse (3342014112155), we get a palindrome (8854128214588).

The spelling of 5512114102433 in words is "five trillion, five hundred twelve billion, one hundred fourteen million, one hundred two thousand, four hundred thirty-three".