Base | Representation |
---|---|
bin | 100101001101100101011… |
… | …0010000110011010010011 |
3 | 200002221012020202120100212 |
4 | 1022123022302012122103 |
5 | 1132243304240311311 |
6 | 14513310132434335 |
7 | 1035334633012562 |
oct | 112331262063223 |
9 | 20087166676325 |
10 | 5114413213331 |
11 | 16a2012a21114 |
12 | 6a725a7789ab |
13 | 2b139699c4b6 |
14 | 139778b23bd9 |
15 | 8d086eb2c8b |
hex | 4a6cac86693 |
5114413213331 has 24 divisors (see below), whose sum is σ = 5283356061312. Its totient is φ = 4947912432000.
The previous prime is 5114413213313. The next prime is 5114413213361. The reversal of 5114413213331 is 1333123144115.
It is a de Polignac number, because none of the positive numbers 2k-5114413213331 is a prime.
It is a super-3 number, since 3×51144132133313 (a number of 39 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (5114413213361) by changing a digit.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 567443981 + ... + 567452993.
It is an arithmetic number, because the mean of its divisors is an integer number (220139835888).
Almost surely, 25114413213331 is an apocalyptic number.
5114413213331 is a deficient number, since it is larger than the sum of its proper divisors (168942847981).
5114413213331 is an equidigital number, since it uses as much as digits as its factorization.
5114413213331 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 9963 (or 9812 counting only the distinct ones).
The product of its digits is 12960, while the sum is 32.
Adding to 5114413213331 its reverse (1333123144115), we get a palindrome (6447536357446).
The spelling of 5114413213331 in words is "five trillion, one hundred fourteen billion, four hundred thirteen million, two hundred thirteen thousand, three hundred thirty-one".
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