110001001010083 has 2 divisors, whose sum is σ = 110001001010084. Its totient is φ = 110001001010082.

The previous prime is 110001001010071. The next prime is 110001001010099. The reversal of 110001001010083 is 380010100100011.

It is an a-pointer prime, because the next prime (110001001010099) can be obtained adding 110001001010083 to its sum of digits (16).

It is a weak prime.

It is an emirp because it is prime and its reverse (380010100100011) is a distict prime.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-110001001010083 is a prime.

It is not a weakly prime, because it can be changed into another prime (110001001010003) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 55000500505041 + 55000500505042.

It is an arithmetic number, because the mean of its divisors is an integer number (55000500505042).

Almost surely, 2^{110001001010083} is an apocalyptic number.

110001001010083 is a deficient number, since it is larger than the sum of its proper divisors (1).

110001001010083 is an equidigital number, since it uses as much as digits as its factorization.

110001001010083 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 24, while the sum is 16.

Adding to 110001001010083 its reverse (380010100100011), we get a palindrome (490011101110094).

The spelling of 110001001010083 in words is "one hundred ten trillion, one billion, one million, ten thousand, eighty-three".