Base | Representation |
---|---|
bin | 100101001110110100011… |
… | …1110001100110101000001 |
3 | 200010012001012200022201101 |
4 | 1022131220332030311001 |
5 | 1132314223144413343 |
6 | 14514425340334401 |
7 | 1035460460263645 |
oct | 112355076146501 |
9 | 20105035608641 |
10 | 5117067185473 |
11 | 16a315502a721 |
12 | 6a787b544401 |
13 | 2b16ca78b809 |
14 | 13994b3b4c25 |
15 | 8d18eea4e4d |
hex | 4a768f8cd41 |
5117067185473 has 2 divisors, whose sum is σ = 5117067185474. Its totient is φ = 5117067185472.
The previous prime is 5117067185453. The next prime is 5117067185491. The reversal of 5117067185473 is 3745817607115.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 4735528720129 + 381538465344 = 2176127^2 + 617688^2 .
It is a cyclic number.
It is not a de Polignac number, because 5117067185473 - 213 = 5117067177281 is a prime.
It is not a weakly prime, because it can be changed into another prime (5117067185423) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2558533592736 + 2558533592737.
It is an arithmetic number, because the mean of its divisors is an integer number (2558533592737).
Almost surely, 25117067185473 is an apocalyptic number.
It is an amenable number.
5117067185473 is a deficient number, since it is larger than the sum of its proper divisors (1).
5117067185473 is an equidigital number, since it uses as much as digits as its factorization.
5117067185473 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 4939200, while the sum is 55.
The spelling of 5117067185473 in words is "five trillion, one hundred seventeen billion, sixty-seven million, one hundred eighty-five thousand, four hundred seventy-three".
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