Base | Representation |
---|---|
bin | 1011101001011101011001… |
… | …0001110110100011000111 |
3 | 1200100022211100200210210002 |
4 | 2322113112101312203013 |
5 | 3134312003440412404 |
6 | 43123223304551515 |
7 | 2461160510141021 |
oct | 272272621664307 |
9 | 50308740623702 |
10 | 12806892513479 |
11 | 4098410a48a29 |
12 | 152a08a48259b |
13 | 71b8b9609bac |
14 | 323bdd4c1011 |
15 | 17320b398c1e |
hex | ba5d64768c7 |
12806892513479 has 2 divisors, whose sum is σ = 12806892513480. Its totient is φ = 12806892513478.
The previous prime is 12806892513469. The next prime is 12806892513553. The reversal of 12806892513479 is 97431529860821.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 12806892513479 - 28 = 12806892513223 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (12806892513469) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6403446256739 + 6403446256740.
It is an arithmetic number, because the mean of its divisors is an integer number (6403446256740).
It is a 1-persistent number, because it is pandigital, but 2⋅12806892513479 = 25613785026958 is not.
Almost surely, 212806892513479 is an apocalyptic number.
12806892513479 is a deficient number, since it is larger than the sum of its proper divisors (1).
12806892513479 is an equidigital number, since it uses as much as digits as its factorization.
12806892513479 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 52254720, while the sum is 65.
The spelling of 12806892513479 in words is "twelve trillion, eight hundred six billion, eight hundred ninety-two million, five hundred thirteen thousand, four hundred seventy-nine".
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