4012500113 has 4 divisors (see below), whose sum is σ = 4120946100. Its totient is φ = 3904054128.

The previous prime is 4012500071. The next prime is 4012500127. The reversal of 4012500113 is 3110052104.

It is a semiprime because it is the product of two primes.

It can be written as a sum of positive squares in 2 ways, for example, as 2473271824 + 1539228289 = 49732^2 + 39233^2 .

It is a cyclic number.

It is not a de Polignac number, because 4012500113 - 2¹² = 4012496017 is a prime.

It is a super-2 number, since 2×4012500113² = 32200314313650025538, which contains 22 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 4012500091 and 4012500100.

It is not an unprimeable number, because it can be changed into a prime (4012500013) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 54222938 + ... + 54223011.

It is an arithmetic number, because the mean of its divisors is an integer number (1030236525).

Almost surely, 2^4012500113 is an apocalyptic number.

It is an amenable number.

4012500113 is a deficient number, since it is larger than the sum of its proper divisors (108445987).

4012500113 is a wasteful number, since it uses less digits than its factorization.

4012500113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 108445986.

The product of its (nonzero) digits is 120, while the sum is 17.

The square root of 4012500113 is about 63344.2981885505. The cubic root of 4012500113 is about 1589.0528901914.

Adding to 4012500113 its reverse (3110052104), we get a palindrome (7122552217).

The spelling of 4012500113 in words is "four billion, twelve million, five hundred thousand, one hundred thirteen".