Base | Representation |
---|---|
bin | 1100000000011001101101… |
… | …0011101111100000111011 |
3 | 1201202000012012000000201011 |
4 | 3000012123103233200323 |
5 | 3212241212234110233 |
6 | 44024250224334351 |
7 | 2531513036545444 |
oct | 300063323574073 |
9 | 51660165000634 |
10 | 13201040144443 |
11 | 422a5910138a8 |
12 | 15925497743b7 |
13 | 749b1162bc78 |
14 | 338d103042cb |
15 | 17d5c90bc0cd |
hex | c019b4ef83b |
13201040144443 has 2 divisors, whose sum is σ = 13201040144444. Its totient is φ = 13201040144442.
The previous prime is 13201040144441. The next prime is 13201040144471. The reversal of 13201040144443 is 34444104010231.
13201040144443 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 is a cyclic number.
It is not a de Polignac number, because 13201040144443 - 21 = 13201040144441 is a prime.
Together with 13201040144441, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (13201040144441) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6600520072221 + 6600520072222.
It is an arithmetic number, because the mean of its divisors is an integer number (6600520072222).
Almost surely, 213201040144443 is an apocalyptic number.
13201040144443 is a deficient number, since it is larger than the sum of its proper divisors (1).
13201040144443 is an equidigital number, since it uses as much as digits as its factorization.
13201040144443 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 18432, while the sum is 31.
Adding to 13201040144443 its reverse (34444104010231), we get a palindrome (47645144154674).
The spelling of 13201040144443 in words is "thirteen trillion, two hundred one billion, forty million, one hundred forty-four thousand, four hundred forty-three".
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