Base | Representation |
---|---|
bin | 101110101011110011… |
… | …111001000001101001 |
3 | 11210101110001220200012 |
4 | 232223303321001221 |
5 | 1310130022201423 |
6 | 35010022421305 |
7 | 3423124502021 |
oct | 565363710151 |
9 | 153343056605 |
10 | 50127147113 |
11 | 1a29351660a |
12 | 986b591835 |
13 | 495b1c9ac4 |
14 | 25d758aa81 |
15 | 1485b0d078 |
hex | babcf9069 |
50127147113 has 2 divisors, whose sum is σ = 50127147114. Its totient is φ = 50127147112.
The previous prime is 50127147107. The next prime is 50127147127. The reversal of 50127147113 is 31174172105.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 44862628864 + 5264518249 = 211808^2 + 72557^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-50127147113 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 50127147113.
It is not a weakly prime, because it can be changed into another prime (50127147173) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25063573556 + 25063573557.
It is an arithmetic number, because the mean of its divisors is an integer number (25063573557).
Almost surely, 250127147113 is an apocalyptic number.
It is an amenable number.
50127147113 is a deficient number, since it is larger than the sum of its proper divisors (1).
50127147113 is an equidigital number, since it uses as much as digits as its factorization.
50127147113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5880, while the sum is 32.
The spelling of 50127147113 in words is "fifty billion, one hundred twenty-seven million, one hundred forty-seven thousand, one hundred thirteen".
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