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13646302664147 is a prime number
BaseRepresentation
bin1100011010010100011100…
…0001011101010111010011
31210022120111021222122002122
43012211013001131113103
53242040111440223042
645005005213404455
72605625046532346
oct306450701352723
953276437878078
1013646302664147
11439140110617a
1216448b319972b
1377cac138553a
143526ad95045d
15189e894ac7d2
hexc694705d5d3

13646302664147 has 2 divisors, whose sum is σ = 13646302664148. Its totient is φ = 13646302664146.

The previous prime is 13646302664131. The next prime is 13646302664161. The reversal of 13646302664147 is 74146620364631.

13646302664147 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 is an emirp because it is prime and its reverse (74146620364631) is a distict prime.

It is a cyclic number.

It is not a de Polignac number, because 13646302664147 - 24 = 13646302664131 is a prime.

It is not a weakly prime, because it can be changed into another prime (13646302664107) by changing a digit.

It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6823151332073 + 6823151332074.

It is an arithmetic number, because the mean of its divisors is an integer number (6823151332074).

Almost surely, 213646302664147 is an apocalyptic number.

13646302664147 is a deficient number, since it is larger than the sum of its proper divisors (1).

13646302664147 is an equidigital number, since it uses as much as digits as its factorization.

13646302664147 is an evil number, because the sum of its binary digits is even.

The product of its (nonzero) digits is 10450944, while the sum is 53.

The spelling of 13646302664147 in words is "thirteen trillion, six hundred forty-six billion, three hundred two million, six hundred sixty-four thousand, one hundred forty-seven".