243120000113 has 4 divisors (see below), whose sum is σ = 248773953648. Its totient is φ = 237466046580.

The previous prime is 243120000107. The next prime is 243120000127. The reversal of 243120000113 is 311000021342.

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.

It is a cyclic number.

It is a de Polignac number, because none of the positive numbers 2^k-243120000113 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 243120000091 and 243120000100.

It is not an unprimeable number, because it can be changed into a prime (243120009113) by changing a digit.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2826976703 + ... + 2826976788.

It is an arithmetic number, because the mean of its divisors is an integer number (62193488412).

Almost surely, 2^243120000113 is an apocalyptic number.

It is an amenable number.

243120000113 is a deficient number, since it is larger than the sum of its proper divisors (5653953535).

243120000113 is an equidigital number, since it uses as much as digits as its factorization.

243120000113 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 5653953534.

The product of its (nonzero) digits is 144, while the sum is 17.

Adding to 243120000113 its reverse (311000021342), we get a palindrome (554120021455).

The spelling of 243120000113 in words is "two hundred forty-three billion, one hundred twenty million, one hundred thirteen", and thus it is an aban number.