142121004305433 has 4 divisors (see below), whose sum is σ = 189494672407248. Its totient is φ = 94747336203620.

The previous prime is 142121004305419. The next prime is 142121004305437. The reversal of 142121004305433 is 334503400121241.

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 142121004305433 - 2¹⁴ = 142121004289049 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 142121004305394 and 142121004305403.

It is not an unprimeable number, because it can be changed into a prime (142121004305437) 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, 23686834050903 + ... + 23686834050908.

It is an arithmetic number, because the mean of its divisors is an integer number (47373668101812).

Almost surely, 2^{142121004305433} is an apocalyptic number.

It is an amenable number.

142121004305433 is a deficient number, since it is larger than the sum of its proper divisors (47373668101815).

142121004305433 is an equidigital number, since it uses as much as digits as its factorization.

142121004305433 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 47373668101814.

The product of its (nonzero) digits is 34560, while the sum is 33.

Adding to 142121004305433 its reverse (334503400121241), we get a palindrome (476624404426674).

The spelling of 142121004305433 in words is "one hundred forty-two trillion, one hundred twenty-one billion, four million, three hundred five thousand, four hundred thirty-three".