12003113 has 4 divisors (see below), whose sum is σ = 12010128. Its totient is φ = 11996100.

The previous prime is 12003109. The next prime is 12003127. The reversal of 12003113 is 31130021.

It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4, and also a brilliant number, because the two primes have the same length.

It is a cyclic number.

It is not a de Polignac number, because 12003113 - 2² = 12003109 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 12003094 and 12003103.

It is not an unprimeable number, because it can be changed into a prime (12003613) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (13) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 938 + ... + 4988.

It is an arithmetic number, because the mean of its divisors is an integer number (3002532).

Almost surely, 2^12003113 is an apocalyptic number.

It is an amenable number.

12003113 is a deficient number, since it is larger than the sum of its proper divisors (7015).

12003113 is an equidigital number, since it uses as much as digits as its factorization.

12003113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 7014.

The product of its (nonzero) digits is 18, while the sum is 11.

The square root of 12003113 is about 3464.5509088481. The cubic root of 12003113 is about 228.9626439959.

Adding to 12003113 its reverse (31130021), we get a palindrome (43133134).

The spelling of 12003113 in words is "twelve million, three thousand, one hundred thirteen".