140122111433 has 2 divisors, whose sum is σ = 140122111434. Its totient is φ = 140122111432.

The previous prime is 140122111421. The next prime is 140122111469. The reversal of 140122111433 is 334111221041.

It is a weak prime.

It can be written as a sum of positive squares in only one way, i.e., 78695958784 + 61426152649 = 280528^2 + 247843^2 .

It is a cyclic number.

It is not a de Polignac number, because 140122111433 - 2⁶ = 140122111369 is a prime.

It is a super-2 number, since 2×140122111433² (a number of 23 digits) contains 22 as substring.

It is a junction number, because it is equal to n+sod(n) for n = 140122111399 and 140122111408.

It is not a weakly prime, because it can be changed into another prime (140122111403) 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, 70061055716 + 70061055717.

It is an arithmetic number, because the mean of its divisors is an integer number (70061055717).

Almost surely, 2^140122111433 is an apocalyptic number.

It is an amenable number.

140122111433 is a deficient number, since it is larger than the sum of its proper divisors (1).

140122111433 is an equidigital number, since it uses as much as digits as its factorization.

140122111433 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 576, while the sum is 23.

Adding to 140122111433 its reverse (334111221041), we get a palindrome (474233332474).

The spelling of 140122111433 in words is "one hundred forty billion, one hundred twenty-two million, one hundred eleven thousand, four hundred thirty-three".