119455443673 has 2 divisors, whose sum is σ = 119455443674. Its totient is φ = 119455443672.

The previous prime is 119455443637. The next prime is 119455443677. The reversal of 119455443673 is 376344554911.

Together with previous prime (119455443637) it forms an Ormiston pair, because they use the same digits, order apart.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 95835442329 + 23620001344 = 309573^2 + 153688^2 .

It is a cyclic number.

It is not a de Polignac number, because 119455443673 - 2⁹ = 119455443161 is a prime.

It is a super-2 number, since 2×119455443673² (a number of 23 digits) contains 22 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 119455443673.

It is not a weakly prime, because it can be changed into another prime (119455443677) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 59727721836 + 59727721837.

It is an arithmetic number, because the mean of its divisors is an integer number (59727721837).

Almost surely, 2^119455443673 is an apocalyptic number.

It is an amenable number.

119455443673 is a deficient number, since it is larger than the sum of its proper divisors (1).

119455443673 is an equidigital number, since it uses as much as digits as its factorization.

119455443673 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 5443200, while the sum is 52.

The spelling of 119455443673 in words is "one hundred nineteen billion, four hundred fifty-five million, four hundred forty-three thousand, six hundred seventy-three".