Base | Representation |
---|---|
bin | 1010111001100001101100… |
… | …0011101100011000111011 |
3 | 1120102121021122011102002111 |
4 | 2232120123003230120323 |
5 | 3032314012123141143 |
6 | 41253034230513151 |
7 | 2344526065244215 |
oct | 256303303543073 |
9 | 46377248142074 |
10 | 11983412708923 |
11 | 3900156081587 |
12 | 14165708057b7 |
13 | 68c05306a656 |
14 | 2d6000a515b5 |
15 | 15bab1abc99d |
hex | ae61b0ec63b |
11983412708923 has 2 divisors, whose sum is σ = 11983412708924. Its totient is φ = 11983412708922.
The previous prime is 11983412708887. The next prime is 11983412708947. The reversal of 11983412708923 is 32980721438911.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 11983412708923 - 213 = 11983412700731 is a prime.
It is not a weakly prime, because it can be changed into another prime (11983412778923) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5991706354461 + 5991706354462.
It is an arithmetic number, because the mean of its divisors is an integer number (5991706354462).
Almost surely, 211983412708923 is an apocalyptic number.
11983412708923 is a deficient number, since it is larger than the sum of its proper divisors (1).
11983412708923 is an equidigital number, since it uses as much as digits as its factorization.
11983412708923 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 5225472, while the sum is 58.
The spelling of 11983412708923 in words is "eleven trillion, nine hundred eighty-three billion, four hundred twelve million, seven hundred eight thousand, nine hundred twenty-three".
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