133203412024433 has 4 divisors (see below), whose sum is σ = 136452275732388. Its totient is φ = 129954548316480.

The previous prime is 133203412024429. The next prime is 133203412024483. The reversal of 133203412024433 is 334420214302331.

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 50029139704384 + 83174272320049 = 7073128^2 + 9119993^2 .

It is a cyclic number.

It is not a de Polignac number, because 133203412024433 - 2² = 133203412024429 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 133203412024393 and 133203412024402.

It is not an unprimeable number, because it can be changed into a prime (133203412024483) 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, 1624431853916 + ... + 1624431853997.

It is an arithmetic number, because the mean of its divisors is an integer number (34113068933097).

Almost surely, 2^{133203412024433} is an apocalyptic number.

It is an amenable number.

133203412024433 is a deficient number, since it is larger than the sum of its proper divisors (3248863707955).

133203412024433 is an equidigital number, since it uses as much as digits as its factorization.

133203412024433 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 3248863707954.

The product of its (nonzero) digits is 124416, while the sum is 35.

Adding to 133203412024433 its reverse (334420214302331), we get a palindrome (467623626326764).

The spelling of 133203412024433 in words is "one hundred thirty-three trillion, two hundred three billion, four hundred twelve million, twenty-four thousand, four hundred thirty-three".