10100313211413 has 4 divisors (see below), whose sum is σ = 13467084281888. Its totient is φ = 6733542140940.

The previous prime is 10100313211379. The next prime is 10100313211447. The reversal of 10100313211413 is 31411231300101.

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 an interprime number because it is at equal distance from previous prime (10100313211379) and next prime (10100313211447).

It is not a de Polignac number, because 10100313211413 - 2¹⁰ = 10100313210389 is a prime.

It is a Duffinian number.

It is a Curzon number.

It is a self number, because there is not a number n which added to its sum of digits gives 10100313211413.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (10100313211493) 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, 1683385535233 + ... + 1683385535238.

It is an arithmetic number, because the mean of its divisors is an integer number (3366771070472).

Almost surely, 2^{10100313211413} is an apocalyptic number.

It is an amenable number.

10100313211413 is a deficient number, since it is larger than the sum of its proper divisors (3366771070475).

10100313211413 is an equidigital number, since it uses as much as digits as its factorization.

10100313211413 is an evil number, because the sum of its binary digits is even.

The sum of its prime factors is 3366771070474.

The product of its (nonzero) digits is 216, while the sum is 21.

Adding to 10100313211413 its reverse (31411231300101), we get a palindrome (41511544511514).

The spelling of 10100313211413 in words is "ten trillion, one hundred billion, three hundred thirteen million, two hundred eleven thousand, four hundred thirteen".