100131000251 has 2 divisors, whose sum is σ = 100131000252. Its totient is φ = 100131000250.

The previous prime is 100131000209. The next prime is 100131000253. The reversal of 100131000251 is 152000131001.

It is a strong prime.

It is an emirp because it is prime and its reverse (152000131001) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 100131000251 - 2⁶ = 100131000187 is a prime.

It is a super-2 number, since 2×100131000251² (a number of 23 digits) contains 22 as substring.

Together with 100131000253, 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 (100131000253) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 50065500125 + 50065500126.

It is an arithmetic number, because the mean of its divisors is an integer number (50065500126).

Almost surely, 2^100131000251 is an apocalyptic number.

100131000251 is a deficient number, since it is larger than the sum of its proper divisors (1).

100131000251 is an equidigital number, since it uses as much as digits as its factorization.

100131000251 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 30, while the sum is 14.

Adding to 100131000251 its reverse (152000131001), we get a palindrome (252131131252).

The spelling of 100131000251 in words is "one hundred billion, one hundred thirty-one million, two hundred fifty-one", and thus it is an aban number.