Base | Representation |
---|---|
bin | 10010101101000010100111… |
… | …11000110111101110001011 |
3 | 12101121222202102002100010122 |
4 | 21112201103320313232023 |
5 | 20342333232121244343 |
6 | 223250511231133455 |
7 | 11443355044453016 |
oct | 1126412370675613 |
9 | 171558672070118 |
10 | 41130014243723 |
11 | 12118160868669 |
12 | 474333267428b |
13 | 19c470b684162 |
14 | a229bbc5837d |
15 | 4b4d456ac068 |
hex | 256853e37b8b |
41130014243723 has 2 divisors, whose sum is σ = 41130014243724. Its totient is φ = 41130014243722.
The previous prime is 41130014243663. The next prime is 41130014243743. The reversal of 41130014243723 is 32734241003114.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 41130014243723 - 224 = 41129997466507 is a prime.
It is a super-3 number, since 3×411300142437233 (a number of 42 digits) contains 333 as substring.
It is a Sophie Germain prime.
It is not a weakly prime, because it can be changed into another prime (41130014243743) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 20565007121861 + 20565007121862.
It is an arithmetic number, because the mean of its divisors is an integer number (20565007121862).
Almost surely, 241130014243723 is an apocalyptic number.
41130014243723 is a deficient number, since it is larger than the sum of its proper divisors (1).
41130014243723 is an equidigital number, since it uses as much as digits as its factorization.
41130014243723 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 48384, while the sum is 35.
Adding to 41130014243723 its reverse (32734241003114), we get a palindrome (73864255246837).
The spelling of 41130014243723 in words is "forty-one trillion, one hundred thirty billion, fourteen million, two hundred forty-three thousand, seven hundred twenty-three".
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