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21102012131 is a prime number
BaseRepresentation
bin10011101001110001…
…110010011011100011
32000110122011021200212
4103221301302123203
5321204103342011
613405533204335
71344633100154
oct235161623343
960418137625
1021102012131
118a49599052
12410b0306ab
131cb3ab1415
1410427c992b
1583789d08b
hex4e9c726e3

21102012131 has 2 divisors, whose sum is σ = 21102012132. Its totient is φ = 21102012130.

The previous prime is 21102012037. The next prime is 21102012133. The reversal of 21102012131 is 13121020112.

It is a strong prime.

It is a cyclic number.

It is not a de Polignac number, because 21102012131 - 230 = 20028270307 is a prime.

It is a super-2 number, since 2×211020121312 (a number of 21 digits) contains 22 as substring.

It is a Sophie Germain prime.

Together with 21102012133, 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 (21102012133) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (19) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10551006065 + 10551006066.

It is an arithmetic number, because the mean of its divisors is an integer number (10551006066).

Almost surely, 221102012131 is an apocalyptic number.

21102012131 is a deficient number, since it is larger than the sum of its proper divisors (1).

21102012131 is an equidigital number, since it uses as much as digits as its factorization.

21102012131 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 24, while the sum is 14.

Adding to 21102012131 its reverse (13121020112), we get a palindrome (34223032243).

The spelling of 21102012131 in words is "twenty-one billion, one hundred two million, twelve thousand, one hundred thirty-one".