Base | Representation |
---|---|
bin | 11101010001000001011… |
… | …111010001101011101001 |
3 | 21010021010001010212002222 |
4 | 131101001133101223221 |
5 | 230422310203144312 |
6 | 4135523451003425 |
7 | 265205002643024 |
oct | 35210137215351 |
9 | 7107101125088 |
10 | 2011143412457 |
11 | 705a15031573 |
12 | 285933a39575 |
13 | 11785b0c235a |
14 | 6d4a88121bb |
15 | 374ab449672 |
hex | 1d4417d1ae9 |
2011143412457 has 2 divisors, whose sum is σ = 2011143412458. Its totient is φ = 2011143412456.
The previous prime is 2011143412433. The next prime is 2011143412469. The reversal of 2011143412457 is 7542143411102.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 1440470438416 + 570672974041 = 1200196^2 + 755429^2 .
It is a cyclic number.
It is not a de Polignac number, because 2011143412457 - 26 = 2011143412393 is a prime.
It is not a weakly prime, because it can be changed into another prime (2011143412417) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 1005571706228 + 1005571706229.
It is an arithmetic number, because the mean of its divisors is an integer number (1005571706229).
Almost surely, 22011143412457 is an apocalyptic number.
It is an amenable number.
2011143412457 is a deficient number, since it is larger than the sum of its proper divisors (1).
2011143412457 is an equidigital number, since it uses as much as digits as its factorization.
2011143412457 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 26880, while the sum is 35.
Adding to 2011143412457 its reverse (7542143411102), we get a palindrome (9553286823559).
The spelling of 2011143412457 in words is "two trillion, eleven billion, one hundred forty-three million, four hundred twelve thousand, four hundred fifty-seven".
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