11331433101401 has 4 divisors (see below), whose sum is σ = 12950209258752. Its totient is φ = 9712656944052.

The previous prime is 11331433101347. The next prime is 11331433101409. The reversal of 11331433101401 is 10410133413311.

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 11331433101401 - 2²⁶ = 11331365992537 is a prime.

It is a super-2 number, since 2×11331433101401² (a number of 27 digits) contains 22 as substring.

It is a Duffinian number.

It is not an unprimeable number, because it can be changed into a prime (11331433101409) 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, 809388078665 + ... + 809388078678.

It is an arithmetic number, because the mean of its divisors is an integer number (3237552314688).

Almost surely, 2^{11331433101401} is an apocalyptic number.

It is an amenable number.

11331433101401 is a deficient number, since it is larger than the sum of its proper divisors (1618776157351).

11331433101401 is an equidigital number, since it uses as much as digits as its factorization.

11331433101401 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 1618776157350.

The product of its (nonzero) digits is 1296, while the sum is 26.

Adding to 11331433101401 its reverse (10410133413311), we get a palindrome (21741566514712).

The spelling of 11331433101401 in words is "eleven trillion, three hundred thirty-one billion, four hundred thirty-three million, one hundred one thousand, four hundred one".