Base | Representation |
---|---|
bin | 1011110100110101010101… |
… | …1110101011011010011001 |
3 | 1201001000012122120020200212 |
4 | 2331031111132223122121 |
5 | 3201012202312013424 |
6 | 43353101343540505 |
7 | 2511246055253024 |
oct | 275152536533231 |
9 | 51030178506625 |
10 | 13002300110489 |
11 | 4163276496858 |
12 | 155bb23bba735 |
13 | 73415b36b941 |
14 | 32d4596c21bb |
15 | 17834657b10e |
hex | bd3557ab699 |
13002300110489 has 2 divisors, whose sum is σ = 13002300110490. Its totient is φ = 13002300110488.
The previous prime is 13002300110399. The next prime is 13002300110531. The reversal of 13002300110489 is 98401100320031.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 12593094255625 + 409205854864 = 3548675^2 + 639692^2 .
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13002300110489 is a prime.
It is not a weakly prime, because it can be changed into another prime (13002300110089) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6501150055244 + 6501150055245.
It is an arithmetic number, because the mean of its divisors is an integer number (6501150055245).
Almost surely, 213002300110489 is an apocalyptic number.
It is an amenable number.
13002300110489 is a deficient number, since it is larger than the sum of its proper divisors (1).
13002300110489 is an equidigital number, since it uses as much as digits as its factorization.
13002300110489 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5184, while the sum is 32.
The spelling of 13002300110489 in words is "thirteen trillion, two billion, three hundred million, one hundred ten thousand, four hundred eighty-nine".
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