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64934023123213 is a prime number
BaseRepresentation
bin11101100001110101000010…
…11000011101000100001101
322111220122001201101001001021
432300322201120131010031
532002334334404420323
6350034140045314141
716451220424462246
oct1660724130350415
9274818051331037
1064934023123213
1119765401254612
127348794b65951
132a30337b5469a
141206b75919ccd
15779139ea505d
hex3b0ea161d10d

64934023123213 has 2 divisors, whose sum is σ = 64934023123214. Its totient is φ = 64934023123212.

The previous prime is 64934023123139. The next prime is 64934023123219. The reversal of 64934023123213 is 31232132043946.

It is a strong prime.

It can be written as a sum of positive squares in only one way, i.e., 64541347940169 + 392675183044 = 8033763^2 + 626638^2 .

It is a cyclic number.

It is not a de Polignac number, because 64934023123213 - 213 = 64934023115021 is a prime.

It is a super-3 number, since 3×649340231232133 (a number of 42 digits) contains 333 as substring.

It is a self number, because there is not a number n which added to its sum of digits gives 64934023123213.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (64934023123219) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 32467011561606 + 32467011561607.

It is an arithmetic number, because the mean of its divisors is an integer number (32467011561607).

Almost surely, 264934023123213 is an apocalyptic number.

It is an amenable number.

64934023123213 is a deficient number, since it is larger than the sum of its proper divisors (1).

64934023123213 is an equidigital number, since it uses as much as digits as its factorization.

64934023123213 is an odious number, because the sum of its binary digits is odd.

The product of its (nonzero) digits is 559872, while the sum is 43.

The spelling of 64934023123213 in words is "sixty-four trillion, nine hundred thirty-four billion, twenty-three million, one hundred twenty-three thousand, two hundred thirteen".