1323102113 has 4 divisors (see below), whose sum is σ = 1323194160. Its totient is φ = 1323010068.

The previous prime is 1323102083. The next prime is 1323102127. The reversal of 1323102113 is 3112013231.

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, and also an emirpimes, since its reverse is a distinct semiprime: 3112013231 = 461 ⋅6750571.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-1323102113 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 1323102091 and 1323102100.

It is not an unprimeable number, because it can be changed into a prime (1323102173) 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 in 3 ways as a sum of consecutive naturals, for example, 19283 + ... + 54936.

It is an arithmetic number, because the mean of its divisors is an integer number (330798540).

Almost surely, 2^1323102113 is an apocalyptic number.

It is an amenable number.

1323102113 is a deficient number, since it is larger than the sum of its proper divisors (92047).

1323102113 is an equidigital number, since it uses as much as digits as its factorization.

1323102113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 92046.

The product of its (nonzero) digits is 108, while the sum is 17.

The square root of 1323102113 is about 36374.4706215774. The cubic root of 1323102113 is about 1097.8199558694.

Adding to 1323102113 its reverse (3112013231), we get a palindrome (4435115344).

The spelling of 1323102113 in words is "one billion, three hundred twenty-three million, one hundred two thousand, one hundred thirteen".