Base | Representation |
---|---|
bin | 10111010000111101111101… |
… | …00110111111010000010111 |
3 | 20201010220022221001010222011 |
4 | 23220132332212333100113 |
5 | 23201203303124314023 |
6 | 300450500031245051 |
7 | 13530141130235113 |
oct | 1350367646772027 |
9 | 221126287033864 |
10 | 51160553354263 |
11 | 15335083685411 |
12 | 58a3311979787 |
13 | 227155b7a1592 |
14 | c8c280c48743 |
15 | 5dac0b71a50d |
hex | 2e87be9bf417 |
51160553354263 has 2 divisors, whose sum is σ = 51160553354264. Its totient is φ = 51160553354262.
The previous prime is 51160553354249. The next prime is 51160553354327. The reversal of 51160553354263 is 36245335506115.
It is a weak prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-51160553354263 is a prime.
It is a super-3 number, since 3×511605533542633 (a number of 42 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (51160553354233) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25580276677131 + 25580276677132.
It is an arithmetic number, because the mean of its divisors is an integer number (25580276677132).
Almost surely, 251160553354263 is an apocalyptic number.
51160553354263 is a deficient number, since it is larger than the sum of its proper divisors (1).
51160553354263 is an equidigital number, since it uses as much as digits as its factorization.
51160553354263 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 4860000, while the sum is 49.
The spelling of 51160553354263 in words is "fifty-one trillion, one hundred sixty billion, five hundred fifty-three million, three hundred fifty-four thousand, two hundred sixty-three".
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