12100211487851 has 2 divisors, whose sum is σ = 12100211487852. Its totient is φ = 12100211487850.

The previous prime is 12100211487829. The next prime is 12100211487853. The reversal of 12100211487851 is 15878411200121.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 12100211487851 - 2²² = 12100207293547 is a prime.

It is a super-2 number, since 2×12100211487851² (a number of 27 digits) contains 22 as substring.

Together with 12100211487853, it forms a pair of twin primes.

It is a Chen prime.

It is a junction number, because it is equal to n+sod(n) for n = 12100211487799 and 12100211487808.

It is not a weakly prime, because it can be changed into another prime (12100211487853) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6050105743925 + 6050105743926.

It is an arithmetic number, because the mean of its divisors is an integer number (6050105743926).

Almost surely, 2^{12100211487851} is an apocalyptic number.

12100211487851 is a deficient number, since it is larger than the sum of its proper divisors (1).

12100211487851 is an equidigital number, since it uses as much as digits as its factorization.

12100211487851 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 35840, while the sum is 41.

Adding to 12100211487851 its reverse (15878411200121), we get a palindrome (27978622687972).

The spelling of 12100211487851 in words is "twelve trillion, one hundred billion, two hundred eleven million, four hundred eighty-seven thousand, eight hundred fifty-one".