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16513806525133 is a prime number
BaseRepresentation
bin1111000001001110101110…
…0100100110001011001101
32011110201000120101122220211
43300103223210212023031
54131030233432301013
655042201001042421
73323040343304041
oct360235344461315
964421016348824
1016513806525133
1152975142a6615
121a285a0486411
1392a325b33988
144113b5a54021
151d9866b7623d
hexf04eb9262cd

16513806525133 has 2 divisors, whose sum is σ = 16513806525134. Its totient is φ = 16513806525132.

The previous prime is 16513806525109. The next prime is 16513806525149. The reversal of 16513806525133 is 33152560831561.

16513806525133 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., 16276205846884 + 237600678249 = 4034378^2 + 487443^2 .

It is an emirp because it is prime and its reverse (33152560831561) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 16513806525133 - 217 = 16513806394061 is a prime.

It is a congruent number.

It is not a weakly prime, because it can be changed into another prime (16513806523133) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 8256903262566 + 8256903262567.

It is an arithmetic number, because the mean of its divisors is an integer number (8256903262567).

Almost surely, 216513806525133 is an apocalyptic number.

It is an amenable number.

16513806525133 is a deficient number, since it is larger than the sum of its proper divisors (1).

16513806525133 is an equidigital number, since it uses as much as digits as its factorization.

16513806525133 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 1944000, while the sum is 49.

The spelling of 16513806525133 in words is "sixteen trillion, five hundred thirteen billion, eight hundred six million, five hundred twenty-five thousand, one hundred thirty-three".