Base | Representation |
---|---|
bin | 1011011000011111001000… |
… | …0001001110000010111101 |
3 | 1122022110011210012102022002 |
4 | 2312013302001032002331 |
5 | 3120022313313423242 |
6 | 42341241032232045 |
7 | 2431125542640602 |
oct | 266076201160275 |
9 | 48273153172262 |
10 | 12515300139197 |
11 | 3a95788723339 |
12 | 14a167083b625 |
13 | 6ca25960295b |
14 | 313a5ac346a9 |
15 | 16a841da3432 |
hex | b61f204e0bd |
12515300139197 has 2 divisors, whose sum is σ = 12515300139198. Its totient is φ = 12515300139196.
The previous prime is 12515300139157. The next prime is 12515300139217. The reversal of 12515300139197 is 79193100351521.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 7106414997796 + 5408885141401 = 2665786^2 + 2325701^2 .
It is a cyclic number.
It is not a de Polignac number, because 12515300139197 - 214 = 12515300122813 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12515300139157) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6257650069598 + 6257650069599.
It is an arithmetic number, because the mean of its divisors is an integer number (6257650069599).
Almost surely, 212515300139197 is an apocalyptic number.
It is an amenable number.
12515300139197 is a deficient number, since it is larger than the sum of its proper divisors (1).
12515300139197 is an equidigital number, since it uses as much as digits as its factorization.
12515300139197 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 255150, while the sum is 47.
The spelling of 12515300139197 in words is "twelve trillion, five hundred fifteen billion, three hundred million, one hundred thirty-nine thousand, one hundred ninety-seven".
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