100312140000131 has 2 divisors, whose sum is σ = 100312140000132. Its totient is φ = 100312140000130.

The previous prime is 100312140000119. The next prime is 100312140000133. The reversal of 100312140000131 is 131000041213001.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 100312140000131 - 2³⁴ = 100294960130947 is a prime.

It is a super-2 number, since 2×100312140000131² (a number of 29 digits) contains 22 as substring.

Together with 100312140000133, it forms a pair of twin primes.

It is a Chen prime.

It is not a weakly prime, because it can be changed into another prime (100312140000133) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50156070000065 + 50156070000066.

It is an arithmetic number, because the mean of its divisors is an integer number (50156070000066).

Almost surely, 2^{100312140000131} is an apocalyptic number.

100312140000131 is a deficient number, since it is larger than the sum of its proper divisors (1).

100312140000131 is an equidigital number, since it uses as much as digits as its factorization.

100312140000131 is an evil number, because the sum of its binary digits is even.

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

Adding to 100312140000131 its reverse (131000041213001), we get a palindrome (231312181213132).

The spelling of 100312140000131 in words is "one hundred trillion, three hundred twelve billion, one hundred forty million, one hundred thirty-one", and thus it is an aban number.