Base | Representation |
---|---|
bin | 1000000000001110100111001… |
… | …1110010000011010011110001 |
3 | 2201212010111020022202212211002 |
4 | 2000003221303302003103301 |
5 | 1042304441124012300243 |
6 | 5313454545221331345 |
7 | 226425633623401301 |
oct | 20003516362032361 |
9 | 2655114208685732 |
10 | 563201004025073 |
11 | 1534a9228174737 |
12 | 53200164866b55 |
13 | 1b23481c0aa4cb |
14 | 9d11107245601 |
15 | 451a23d392eb8 |
hex | 2003a73c834f1 |
563201004025073 has 2 divisors, whose sum is σ = 563201004025074. Its totient is φ = 563201004025072.
The previous prime is 563201004025031. The next prime is 563201004025103. The reversal of 563201004025073 is 370520400102365.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 405037883555809 + 158163120469264 = 20125553^2 + 12576292^2 .
It is a cyclic number.
It is not a de Polignac number, because 563201004025073 - 210 = 563201004024049 is a prime.
It is not a weakly prime, because it can be changed into another prime (563201004025013) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 281600502012536 + 281600502012537.
It is an arithmetic number, because the mean of its divisors is an integer number (281600502012537).
Almost surely, 2563201004025073 is an apocalyptic number.
It is an amenable number.
563201004025073 is a deficient number, since it is larger than the sum of its proper divisors (1).
563201004025073 is an equidigital number, since it uses as much as digits as its factorization.
563201004025073 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 151200, while the sum is 38.
The spelling of 563201004025073 in words is "five hundred sixty-three trillion, two hundred one billion, four million, twenty-five thousand, seventy-three".
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