Base | Representation |
---|---|
bin | 11010011001011010000000… |
… | …111101000001011010101001 |
3 | 120020001120211012000012002222 |
4 | 122121122000331001122221 |
5 | 110204100311122203423 |
6 | 1050525211034003425 |
7 | 33311411351413133 |
oct | 3231320075013251 |
9 | 506046735005088 |
10 | 116095129491113 |
11 | 33a9a739005293 |
12 | 1103001bb76575 |
13 | 4ca1954080066 |
14 | 2095069d9a053 |
15 | d64d7a00b3c8 |
hex | 699680f416a9 |
116095129491113 has 2 divisors, whose sum is σ = 116095129491114. Its totient is φ = 116095129491112.
The previous prime is 116095129491101. The next prime is 116095129491193. The reversal of 116095129491113 is 311194921590611.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 115706960618089 + 388168873024 = 10756717^2 + 623032^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-116095129491113 is a prime.
It is not a weakly prime, because it can be changed into another prime (116095129491193) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 58047564745556 + 58047564745557.
It is an arithmetic number, because the mean of its divisors is an integer number (58047564745557).
Almost surely, 2116095129491113 is an apocalyptic number.
It is an amenable number.
116095129491113 is a deficient number, since it is larger than the sum of its proper divisors (1).
116095129491113 is an equidigital number, since it uses as much as digits as its factorization.
116095129491113 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 524880, while the sum is 53.
The spelling of 116095129491113 in words is "one hundred sixteen trillion, ninety-five billion, one hundred twenty-nine million, four hundred ninety-one thousand, one hundred thirteen".
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