Base | Representation |
---|---|
bin | 10011000101000100100… |
… | …110110101001100001011 |
3 | 11122100012221011120102021 |
4 | 103011010212311030023 |
5 | 132440131202232212 |
6 | 2442152435145311 |
7 | 163503451406032 |
oct | 23050446651413 |
9 | 4570187146367 |
10 | 1311116055307 |
11 | 46604a918063 |
12 | 19212a145237 |
13 | 9683a097366 |
14 | 4765ba29719 |
15 | 24189c8b307 |
hex | 131449b530b |
1311116055307 has 2 divisors, whose sum is σ = 1311116055308. Its totient is φ = 1311116055306.
The previous prime is 1311116055251. The next prime is 1311116055421. The reversal of 1311116055307 is 7035506111131.
It is a weak prime.
It is an emirp because it is prime and its reverse (7035506111131) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1311116055307 - 215 = 1311116022539 is a prime.
It is not a weakly prime, because it can be changed into another prime (1311116055607) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 655558027653 + 655558027654.
It is an arithmetic number, because the mean of its divisors is an integer number (655558027654).
Almost surely, 21311116055307 is an apocalyptic number.
1311116055307 is a deficient number, since it is larger than the sum of its proper divisors (1).
1311116055307 is an equidigital number, since it uses as much as digits as its factorization.
1311116055307 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 9450, while the sum is 34.
The spelling of 1311116055307 in words is "one trillion, three hundred eleven billion, one hundred sixteen million, fifty-five thousand, three hundred seven".
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