10300125113 has 4 divisors (see below), whose sum is σ = 10305643920. Its totient is φ = 10294606308.

The previous prime is 10300125109. The next prime is 10300125133. The reversal of 10300125113 is 31152100301.

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 not a de Polignac number, because 10300125113 - 2² = 10300125109 is a prime.

It is a super-3 number, since 3×10300125113³ (a number of 31 digits) contains 333 as substring.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 10300125091 and 10300125100.

It is not an unprimeable number, because it can be changed into a prime (10300125133) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (23) of ones.

It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 2756603 + ... + 2760336.

It is an arithmetic number, because the mean of its divisors is an integer number (2576410980).

Almost surely, 2^10300125113 is an apocalyptic number.

It is an amenable number.

10300125113 is a deficient number, since it is larger than the sum of its proper divisors (5518807).

10300125113 is an equidigital number, since it uses as much as digits as its factorization.

10300125113 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 5518806.

The product of its (nonzero) digits is 90, while the sum is 17.

Adding to 10300125113 its reverse (31152100301), we get a palindrome (41452225414).

The spelling of 10300125113 in words is "ten billion, three hundred million, one hundred twenty-five thousand, one hundred thirteen".