11325501123133 has 2 divisors, whose sum is σ = 11325501123134. Its totient is φ = 11325501123132.

The previous prime is 11325501123121. The next prime is 11325501123221. The reversal of 11325501123133 is 33132110552311.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 11232438299289 + 93062823844 = 3351483^2 + 305062^2 .

It is an emirp because it is prime and its reverse (33132110552311) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-11325501123133 is a prime.

It is a super-3 number, since 3×11325501123133³ (a number of 40 digits) contains 333 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 11325501123095 and 11325501123104.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (11325501123533) 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, 5662750561566 + 5662750561567.

It is an arithmetic number, because the mean of its divisors is an integer number (5662750561567).

Almost surely, 2^{11325501123133} is an apocalyptic number.

It is an amenable number.

11325501123133 is a deficient number, since it is larger than the sum of its proper divisors (1).

11325501123133 is an equidigital number, since it uses as much as digits as its factorization.

11325501123133 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 8100, while the sum is 31.

Adding to 11325501123133 its reverse (33132110552311), we get a palindrome (44457611675444).

The spelling of 11325501123133 in words is "eleven trillion, three hundred twenty-five billion, five hundred one million, one hundred twenty-three thousand, one hundred thirty-three".