Base | Representation |
---|---|
bin | 10111101101011100101110… |
… | …11101001010110010110011 |
3 | 20211121110020221102022222111 |
4 | 23312232113131022302303 |
5 | 23313221434121342042 |
6 | 302520221104350151 |
7 | 13660633501131301 |
oct | 1366562735126263 |
9 | 224543227368874 |
10 | 52139149012147 |
11 | 156820a839726a |
12 | 5a20abb615357 |
13 | 2312915453512 |
14 | cc3796687271 |
15 | 6063d3cb4517 |
hex | 2f6b9774acb3 |
52139149012147 has 2 divisors, whose sum is σ = 52139149012148. Its totient is φ = 52139149012146.
The previous prime is 52139149012103. The next prime is 52139149012163. The reversal of 52139149012147 is 74121094193125.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 52139149012147 - 27 = 52139149012019 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 52139149012147.
It is not a weakly prime, because it can be changed into another prime (52139149015147) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26069574506073 + 26069574506074.
It is an arithmetic number, because the mean of its divisors is an integer number (26069574506074).
Almost surely, 252139149012147 is an apocalyptic number.
52139149012147 is a deficient number, since it is larger than the sum of its proper divisors (1).
52139149012147 is an equidigital number, since it uses as much as digits as its factorization.
52139149012147 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 544320, while the sum is 49.
The spelling of 52139149012147 in words is "fifty-two trillion, one hundred thirty-nine billion, one hundred forty-nine million, twelve thousand, one hundred forty-seven".
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