100101314000213 has 4 divisors (see below), whose sum is σ = 100831980525900. Its totient is φ = 99370647474528.

The previous prime is 100101314000197. The next prime is 100101314000327. The reversal of 100101314000213 is 312000413101001.

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 37462209527044 + 62639104473169 = 6120638^2 + 7914487^2 .

It is a cyclic number.

It is not a de Polignac number, because 100101314000213 - 2⁴ = 100101314000197 is a prime.

It is a Duffinian number.

It is a junction number, because it is equal to n+sod(n) for n = 100101314000191 and 100101314000200.

It is a congruent number.

It is not an unprimeable number, because it can be changed into a prime (100101314060213) 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, 365333262638 + ... + 365333262911.

It is an arithmetic number, because the mean of its divisors is an integer number (25207995131475).

Almost surely, 2^{100101314000213} is an apocalyptic number.

It is an amenable number.

100101314000213 is a deficient number, since it is larger than the sum of its proper divisors (730666525687).

100101314000213 is an equidigital number, since it uses as much as digits as its factorization.

100101314000213 is an odious number, because the sum of its binary digits is odd.

The sum of its prime factors is 730666525686.

The product of its (nonzero) digits is 72, while the sum is 17.

Adding to 100101314000213 its reverse (312000413101001), we get a palindrome (412101727101214).

The spelling of 100101314000213 in words is "one hundred trillion, one hundred one billion, three hundred fourteen million, two hundred thirteen", and thus it is an aban number.