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6314524254317 is a prime number
BaseRepresentation
bin101101111100011011011…
…1110011010000001101101
3211100122220001102010221102
41123320312332122001231
51311424121202114232
621232504041212445
71221131540364356
oct133706676320155
924318801363842
106314524254317
112014a7a35aa14
1285b969491125
1336a5c3295316
1417b8a5cb4d2d
15ae3c67d4662
hex5be36f9a06d

6314524254317 has 2 divisors, whose sum is σ = 6314524254318. Its totient is φ = 6314524254316.

The previous prime is 6314524254133. The next prime is 6314524254319. The reversal of 6314524254317 is 7134524254136.

6314524254317 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 4667937412681 + 1646586841636 = 2160541^2 + 1283194^2 .

It is a cyclic number.

It is not a de Polignac number, because 6314524254317 - 214 = 6314524237933 is a prime.

Together with 6314524254319, it forms a pair of twin primes.

It is a Chen prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (6314524254319) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 3157262127158 + 3157262127159.

It is an arithmetic number, because the mean of its divisors is an integer number (3157262127159).

Almost surely, 26314524254317 is an apocalyptic number.

It is an amenable number.

6314524254317 is a deficient number, since it is larger than the sum of its proper divisors (1).

6314524254317 is an equidigital number, since it uses as much as digits as its factorization.

6314524254317 is an odious number, because the sum of its binary digits is odd.

The product of its digits is 2419200, while the sum is 47.

The spelling of 6314524254317 in words is "six trillion, three hundred fourteen billion, five hundred twenty-four million, two hundred fifty-four thousand, three hundred seventeen".