10043350113 has 4 divisors (see below), whose sum is σ = 13391133488. Its totient is φ = 6695566740.

The previous prime is 10043350069. The next prime is 10043350123. The reversal of 10043350113 is 31105334001.

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 not a de Polignac number, because 10043350113 - 2⁹ = 10043349601 is a prime.

It is a super-2 number, since 2×10043350113² (a number of 21 digits) contains 22 as substring.

It is a Duffinian number.

It is a self number, because there is not a number n which added to its sum of digits gives 10043350113.

It is not an unprimeable number, because it can be changed into a prime (10043350123) 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, 1673891683 + ... + 1673891688.

It is an arithmetic number, because the mean of its divisors is an integer number (3347783372).

Almost surely, 2^10043350113 is an apocalyptic number.

It is an amenable number.

10043350113 is a deficient number, since it is larger than the sum of its proper divisors (3347783375).

10043350113 is an equidigital number, since it uses as much as digits as its factorization.

10043350113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 3347783374.

The product of its (nonzero) digits is 540, while the sum is 21.

Adding to 10043350113 its reverse (31105334001), we get a palindrome (41148684114).

The spelling of 10043350113 in words is "ten billion, forty-three million, three hundred fifty thousand, one hundred thirteen".