Base | Representation |
---|---|
bin | 1010111001100001101100… |
… | …0011101100001011001001 |
3 | 1120102121021122011100212211 |
4 | 2232120123003230023021 |
5 | 3032314012123124131 |
6 | 41253034230505121 |
7 | 2344526065241515 |
oct | 256303303541311 |
9 | 46377248140784 |
10 | 11983412708041 |
11 | 3900156080955 |
12 | 14165708051a1 |
13 | 68c05306a128 |
14 | 2d6000a51145 |
15 | 15bab1abc5b1 |
hex | ae61b0ec2c9 |
11983412708041 has 2 divisors, whose sum is σ = 11983412708042. Its totient is φ = 11983412708040.
The previous prime is 11983412707999. The next prime is 11983412708107. The reversal of 11983412708041 is 14080721438911.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 9814567418041 + 2168845290000 = 3132821^2 + 1472700^2 .
It is a cyclic number.
It is not a de Polignac number, because 11983412708041 - 29 = 11983412707529 is a prime.
It is not a weakly prime, because it can be changed into another prime (11983412705041) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5991706354020 + 5991706354021.
It is an arithmetic number, because the mean of its divisors is an integer number (5991706354021).
Almost surely, 211983412708041 is an apocalyptic number.
It is an amenable number.
11983412708041 is a deficient number, since it is larger than the sum of its proper divisors (1).
11983412708041 is an equidigital number, since it uses as much as digits as its factorization.
11983412708041 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 387072, while the sum is 49.
The spelling of 11983412708041 in words is "eleven trillion, nine hundred eighty-three billion, four hundred twelve million, seven hundred eight thousand, forty-one".
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