Base | Representation |
---|---|
bin | 101101011010111101111… |
… | …000101100010000001001 |
3 | 102001101200222212002221212 |
4 | 231122331320230100021 |
5 | 402114441443341131 |
6 | 10345530425342505 |
7 | 441336342225365 |
oct | 55327570542011 |
9 | 12041628762855 |
10 | 3121332012041 |
11 | aa38276031a1 |
12 | 424b27937a35 |
13 | 19845697cc79 |
14 | ab1050d42a5 |
15 | 562d660ea2b |
hex | 2d6bde2c409 |
3121332012041 has 2 divisors, whose sum is σ = 3121332012042. Its totient is φ = 3121332012040.
The previous prime is 3121332012011. The next prime is 3121332012053. The reversal of 3121332012041 is 1402102331213.
3121332012041 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 2953658141641 + 167673870400 = 1718621^2 + 409480^2 .
It is an emirp because it is prime and its reverse (1402102331213) is a distict prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-3121332012041 is a prime.
It is a Sophie Germain prime.
It is a Curzon number.
It is a junction number, because it is equal to n+sod(n) for n = 3121332011998 and 3121332012016.
It is not a weakly prime, because it can be changed into another prime (3121332012011) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1560666006020 + 1560666006021.
It is an arithmetic number, because the mean of its divisors is an integer number (1560666006021).
Almost surely, 23121332012041 is an apocalyptic number.
It is an amenable number.
3121332012041 is a deficient number, since it is larger than the sum of its proper divisors (1).
3121332012041 is an equidigital number, since it uses as much as digits as its factorization.
3121332012041 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 864, while the sum is 23.
Adding to 3121332012041 its reverse (1402102331213), we get a palindrome (4523434343254).
The spelling of 3121332012041 in words is "three trillion, one hundred twenty-one billion, three hundred thirty-two million, twelve thousand, forty-one".
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