Base | Representation |
---|---|
bin | 100110111010001000010101… |
… | …110000110101010100100011 |
3 | 211102212221210112021220221102 |
4 | 212322020111300311110203 |
5 | 134412114141221130313 |
6 | 1403535352255545015 |
7 | 51021021216600251 |
oct | 4672102560652443 |
9 | 742787715256842 |
10 | 171120452130083 |
11 | 4a58485a143335 |
12 | 1723835922b16b |
13 | 7463787b96c21 |
14 | 30383ba0bb5d1 |
15 | 14bb380541a58 |
hex | 9ba215c35523 |
171120452130083 has 2 divisors, whose sum is σ = 171120452130084. Its totient is φ = 171120452130082.
The previous prime is 171120452130031. The next prime is 171120452130193. The reversal of 171120452130083 is 380031254021171.
It is a weak prime.
It is an emirp because it is prime and its reverse (380031254021171) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 171120452130083 - 220 = 171120451081507 is a prime.
It is not a weakly prime, because it can be changed into another prime (171120452133083) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 85560226065041 + 85560226065042.
It is an arithmetic number, because the mean of its divisors is an integer number (85560226065042).
Almost surely, 2171120452130083 is an apocalyptic number.
171120452130083 is a deficient number, since it is larger than the sum of its proper divisors (1).
171120452130083 is an equidigital number, since it uses as much as digits as its factorization.
171120452130083 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 40320, while the sum is 38.
The spelling of 171120452130083 in words is "one hundred seventy-one trillion, one hundred twenty billion, four hundred fifty-two million, one hundred thirty thousand, eighty-three".
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