1211032325141 has 2 divisors, whose sum is σ = 1211032325142. Its totient is φ = 1211032325140.

The previous prime is 1211032325101. The next prime is 1211032325143. The reversal of 1211032325141 is 1415232301121.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 1207833970225 + 3198354916 = 1099015^2 + 56554^2 .

It is an emirp because it is prime and its reverse (1415232301121) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 1211032325141 - 2¹⁸ = 1211032062997 is a prime.

It is a super-3 number, since 3×1211032325141³ (a number of 37 digits) contains 333 as substring.

Together with 1211032325143, it forms a pair of twin primes.

It is a Chen prime.

It is a self number, because there is not a number n which added to its sum of digits gives 1211032325141.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (1211032325143) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 605516162570 + 605516162571.

It is an arithmetic number, because the mean of its divisors is an integer number (605516162571).

Almost surely, 2^{1211032325141} is an apocalyptic number.

It is an amenable number.

1211032325141 is a deficient number, since it is larger than the sum of its proper divisors (1).

1211032325141 is an equidigital number, since it uses as much as digits as its factorization.

1211032325141 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1440, while the sum is 26.

Adding to 1211032325141 its reverse (1415232301121), we get a palindrome (2626264626262).

The spelling of 1211032325141 in words is "one trillion, two hundred eleven billion, thirty-two million, three hundred twenty-five thousand, one hundred forty-one".