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13201064050081 is a prime number
BaseRepresentation
bin1100000000011001110010…
…1110111011110110100001
31201202000020211222120002011
43000012130232323312201
53212241234344100311
644024252440552521
72531513450002255
oct300063456736641
951660224876064
1013201064050081
11422a5a3561526
121592555782741
13749b1657c00b
14338d13568265
1517d5cb240321
hexc019cbbbda1

13201064050081 has 2 divisors, whose sum is σ = 13201064050082. Its totient is φ = 13201064050080.

The previous prime is 13201064050033. The next prime is 13201064050127. The reversal of 13201064050081 is 18005046010231.

13201064050081 is digitally balanced in base 2, 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., 13175266550625 + 25797499456 = 3629775^2 + 160616^2 .

It is a cyclic number.

It is not a de Polignac number, because 13201064050081 - 225 = 13201030495649 is a prime.

It is a super-2 number, since 2×132010640500812 (a number of 27 digits) contains 22 as substring.

It is not a weakly prime, because it can be changed into another prime (13201064050001) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6600532025040 + 6600532025041.

It is an arithmetic number, because the mean of its divisors is an integer number (6600532025041).

Almost surely, 213201064050081 is an apocalyptic number.

It is an amenable number.

13201064050081 is a deficient number, since it is larger than the sum of its proper divisors (1).

13201064050081 is an equidigital number, since it uses as much as digits as its factorization.

13201064050081 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 5760, while the sum is 31.

The spelling of 13201064050081 in words is "thirteen trillion, two hundred one billion, sixty-four million, fifty thousand, eighty-one".