10001119999963 has 2 divisors, whose sum is σ = 10001119999964. Its totient is φ = 10001119999962.

The previous prime is 10001119999961. The next prime is 10001120000161. The reversal of 10001119999963 is 36999991110001.

10001119999963 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is a cyclic number.

It is not a de Polignac number, because 10001119999963 - 2¹ = 10001119999961 is a prime.

Together with 10001119999961, it forms a pair of twin primes.

It is a junction number, because it is equal to n+sod(n) for n = 10001119999898 and 10001119999907.

It is not a weakly prime, because it can be changed into another prime (10001119999961) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5000559999981 + 5000559999982.

It is an arithmetic number, because the mean of its divisors is an integer number (5000559999982).

Almost surely, 2^{10001119999963} is an apocalyptic number.

10001119999963 is a deficient number, since it is larger than the sum of its proper divisors (1).

10001119999963 is an equidigital number, since it uses as much as digits as its factorization.

10001119999963 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1062882, while the sum is 58.

The spelling of 10001119999963 in words is "ten trillion, one billion, one hundred nineteen million, nine hundred ninety-nine thousand, nine hundred sixty-three".