2021114132433 has 64 divisors (see below), whose sum is σ = 3279335731200. Its totient is φ = 1220088070080.

The previous prime is 2021114132419. The next prime is 2021114132447. The reversal of 2021114132433 is 3342314111202.

2021114132433 is a `hidden beast` number, since 2 + 0 + 2 + 1 + 1 + 1 + 413 + 243 + 3 = 666.

It is an interprime number because it is at equal distance from previous prime (2021114132419) and next prime (2021114132447).

It is not a de Polignac number, because 2021114132433 - 2²¹ = 2021112035281 is a prime.

It is a Harshad number since it is a multiple of its sum of digits (27).

It is a junction number, because it is equal to n+sod(n) for n = 2021114132397 and 2021114132406.

It is not an unprimeable number, because it can be changed into a prime (2021114132333) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 63 ways as a sum of consecutive naturals, for example, 366207448 + ... + 366212966.

It is an arithmetic number, because the mean of its divisors is an integer number (51239620800).

Almost surely, 2^{2021114132433} is an apocalyptic number.

It is an amenable number.

2021114132433 is a deficient number, since it is larger than the sum of its proper divisors (1258221598767).

2021114132433 is a wasteful number, since it uses less digits than its factorization.

2021114132433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 10179 (or 10173 counting only the distinct ones).

The product of its (nonzero) digits is 3456, while the sum is 27.

Adding to 2021114132433 its reverse (3342314111202), we get a palindrome (5363428243635).

It can be divided in two parts, 20211 and 14132433, that added together give a square (14152644 = 3762²).

The spelling of 2021114132433 in words is "two trillion, twenty-one billion, one hundred fourteen million, one hundred thirty-two thousand, four hundred thirty-three".