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12433199323 is a prime number
BaseRepresentation
bin10111001010001001…
…11001000011011011
31012002111020222002211
423211010321003123
5200430344334243
65413422335551
7620051310346
oct134504710333
935074228084
1012433199323
115300242746
1224aba2a5b7
131231b2b39c
1485d38545d
154cb7e0d9d
hex2e51390db

12433199323 has 2 divisors, whose sum is σ = 12433199324. Its totient is φ = 12433199322.

The previous prime is 12433199321. The next prime is 12433199371. The reversal of 12433199323 is 32399133421.

12433199323 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.

It is a weak prime.

It is an emirp because it is prime and its reverse (32399133421) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 12433199323 - 21 = 12433199321 is a prime.

It is a super-3 number, since 3×124331993233 (a number of 31 digits) contains 333 as substring.

Together with 12433199321, it forms a pair of twin primes.

It is not a weakly prime, because it can be changed into another prime (12433199321) by changing a digit.

It is a pernicious number, because its binary representation contains a prime number (17) of ones.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6216599661 + 6216599662.

It is an arithmetic number, because the mean of its divisors is an integer number (6216599662).

Almost surely, 212433199323 is an apocalyptic number.

12433199323 is a deficient number, since it is larger than the sum of its proper divisors (1).

12433199323 is an equidigital number, since it uses as much as digits as its factorization.

12433199323 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 104976, while the sum is 40.

The spelling of 12433199323 in words is "twelve billion, four hundred thirty-three million, one hundred ninety-nine thousand, three hundred twenty-three".