Base | Representation |
---|---|
bin | 111100100010010011… |
… | …010100110111100001 |
3 | 20012202221222011200012 |
4 | 330202103110313201 |
5 | 2031110000012433 |
6 | 45505515334305 |
7 | 4460521615034 |
oct | 744223246741 |
9 | 205687864605 |
10 | 65000000993 |
11 | 25625895894 |
12 | 10720451395 |
13 | 618b5b9365 |
14 | 320895501b |
15 | 1a566a9848 |
hex | f224d4de1 |
65000000993 has 2 divisors, whose sum is σ = 65000000994. Its totient is φ = 65000000992.
The previous prime is 65000000987. The next prime is 65000001019. The reversal of 65000000993 is 39900000056.
65000000993 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 61277536849 + 3722464144 = 247543^2 + 61012^2 .
It is a cyclic number.
It is not a de Polignac number, because 65000000993 - 228 = 64731565537 is a prime.
It is not a weakly prime, because it can be changed into another prime (65000000953) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 32500000496 + 32500000497.
It is an arithmetic number, because the mean of its divisors is an integer number (32500000497).
Almost surely, 265000000993 is an apocalyptic number.
It is an amenable number.
65000000993 is a deficient number, since it is larger than the sum of its proper divisors (1).
65000000993 is an equidigital number, since it uses as much as digits as its factorization.
65000000993 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 7290, while the sum is 32.
The spelling of 65000000993 in words is "sixty-five billion, nine hundred ninety-three", and thus it is an aban number.
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