999988000003 has 2 divisors, whose sum is σ = 999988000004. Its totient is φ = 999988000002.

The previous prime is 999987999949. The next prime is 999988000057. The reversal of 999988000003 is 300000889999.

It is a balanced prime because it is at equal distance from previous prime (999987999949) and next prime (999988000057).

It is an emirp because it is prime and its reverse (300000889999) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 999988000003 - 2¹⁷ = 999987868931 is a prime.

It is a junction number, because it is equal to n+sod(n) for n = 999987999899 and 999987999908.

It is not a weakly prime, because it can be changed into another prime (999988000093) 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, 499994000001 + 499994000002.

It is an arithmetic number, because the mean of its divisors is an integer number (499994000002).

Almost surely, 2^999988000003 is an apocalyptic number.

999988000003 is a deficient number, since it is larger than the sum of its proper divisors (1).

999988000003 is an equidigital number, since it uses as much as digits as its factorization.

999988000003 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 1259712, while the sum is 55.

The spelling of 999988000003 in words is "nine hundred ninety-nine billion, nine hundred eighty-eight million, three", and thus it is an aban number.