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12806892513553 is a prime number
BaseRepresentation
bin1011101001011101011001…
…0001110110100100010001
31200100022211100200210212211
42322113112101312210101
53134312003440413203
643123223304552121
72461160510141155
oct272272621664421
950308740623784
1012806892513553
114098410a48a96
12152a08a482641
1371b8b9609c38
14323bdd4c1065
1517320b398c6d
hexba5d6476911

12806892513553 has 2 divisors, whose sum is σ = 12806892513554. Its totient is φ = 12806892513552.

The previous prime is 12806892513479. The next prime is 12806892513557. The reversal of 12806892513553 is 35531529860821.

12806892513553 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., 6981099299329 + 5825793214224 = 2642177^2 + 2413668^2 .

It is a cyclic number.

It is not a de Polignac number, because 12806892513553 - 221 = 12806890416401 is a prime.

It is not a weakly prime, because it can be changed into another prime (12806892513557) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6403446256776 + 6403446256777.

It is an arithmetic number, because the mean of its divisors is an integer number (6403446256777).

Almost surely, 212806892513553 is an apocalyptic number.

It is an amenable number.

12806892513553 is a deficient number, since it is larger than the sum of its proper divisors (1).

12806892513553 is an equidigital number, since it uses as much as digits as its factorization.

12806892513553 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 15552000, while the sum is 58.

The spelling of 12806892513553 in words is "twelve trillion, eight hundred six billion, eight hundred ninety-two million, five hundred thirteen thousand, five hundred fifty-three".