Base | Representation |
---|---|
bin | 11010001011100000100110… |
… | …100011001110100011010001 |
3 | 120002200021210122001111122101 |
4 | 122023200212203032203101 |
5 | 110042423441242240032 |
6 | 1044514351233240401 |
7 | 33152412521126515 |
oct | 3213404643164321 |
9 | 502607718044571 |
10 | 115140130040017 |
11 | 3376172278a885 |
12 | 10ab6b18703101 |
13 | 4c3288ab58464 |
14 | 2060b522c6145 |
15 | d4a0d4290de7 |
hex | 68b8268ce8d1 |
115140130040017 has 2 divisors, whose sum is σ = 115140130040018. Its totient is φ = 115140130040016.
The previous prime is 115140130039973. The next prime is 115140130040021. The reversal of 115140130040017 is 710040031041511.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 106792878756096 + 8347251283921 = 10334064^2 + 2889161^2 .
It is a cyclic number.
It is not a de Polignac number, because 115140130040017 - 215 = 115140130007249 is a prime.
It is not a weakly prime, because it can be changed into another prime (115140130046017) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57570065020008 + 57570065020009.
It is an arithmetic number, because the mean of its divisors is an integer number (57570065020009).
Almost surely, 2115140130040017 is an apocalyptic number.
It is an amenable number.
115140130040017 is a deficient number, since it is larger than the sum of its proper divisors (1).
115140130040017 is an equidigital number, since it uses as much as digits as its factorization.
115140130040017 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1680, while the sum is 28.
Adding to 115140130040017 its reverse (710040031041511), we get a palindrome (825180161081528).
The spelling of 115140130040017 in words is "one hundred fifteen trillion, one hundred forty billion, one hundred thirty million, forty thousand, seventeen".
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