11331132111133 has 2 divisors, whose sum is σ = 11331132111134. Its totient is φ = 11331132111132.

The previous prime is 11331132111109. The next prime is 11331132111157. The reversal of 11331132111133 is 33111123113311.

11331132111133 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a balanced prime because it is at equal distance from previous prime (11331132111109) and next prime (11331132111157).

It can be written as a sum of positive squares in only one way, i.e., 6866354659129 + 4464777452004 = 2620373^2 + 2113002^2 .

It is an emirp because it is prime and its reverse (33111123113311) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-11331132111133 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 11331132111098 and 11331132111107.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (11331132111163) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5665566055566 + 5665566055567.

It is an arithmetic number, because the mean of its divisors is an integer number (5665566055567).

Almost surely, 2^{11331132111133} is an apocalyptic number.

It is an amenable number.

11331132111133 is a deficient number, since it is larger than the sum of its proper divisors (1).

11331132111133 is an equidigital number, since it uses as much as digits as its factorization.

11331132111133 is an evil number, because the sum of its binary digits is even.

The product of its digits is 486, while the sum is 25.

Adding to 11331132111133 its reverse (33111123113311), we get a palindrome (44442255224444).

The spelling of 11331132111133 in words is "eleven trillion, three hundred thirty-one billion, one hundred thirty-two million, one hundred eleven thousand, one hundred thirty-three".