13111400235433 has 2 divisors, whose sum is σ = 13111400235434. Its totient is φ = 13111400235432.

The previous prime is 13111400235397. The next prime is 13111400235467. The reversal of 13111400235433 is 33453200411131.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 8740135402384 + 4371264833049 = 2956372^2 + 2090757^2 .

It is a cyclic number.

It is not a de Polignac number, because 13111400235433 - 2⁹ = 13111400234921 is a prime.

It is a super-2 number, since 2×13111400235433² (a number of 27 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 13111400235395 and 13111400235404.

It is not a weakly prime, because it can be changed into another prime (13111400235233) 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 as a sum of consecutive naturals, namely, 6555700117716 + 6555700117717.

It is an arithmetic number, because the mean of its divisors is an integer number (6555700117717).

Almost surely, 2^{13111400235433} is an apocalyptic number.

It is an amenable number.

13111400235433 is a deficient number, since it is larger than the sum of its proper divisors (1).

13111400235433 is an equidigital number, since it uses as much as digits as its factorization.

13111400235433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 12960, while the sum is 31.

Adding to 13111400235433 its reverse (33453200411131), we get a palindrome (46564600646564).

The spelling of 13111400235433 in words is "thirteen trillion, one hundred eleven billion, four hundred million, two hundred thirty-five thousand, four hundred thirty-three".