Base | Representation |
---|---|
bin | 1100010001111010111010… |
… | …1101100111100101100011 |
3 | 1202210210001112021211200112 |
4 | 3010132232231213211203 |
5 | 3232204110412434124 |
6 | 44414423410041535 |
7 | 2562326323215041 |
oct | 304365655474543 |
9 | 52723045254615 |
10 | 13502013405539 |
11 | 433619863060a |
12 | 1620944b792ab |
13 | 76c307c951c9 |
14 | 3497027d3391 |
15 | 186341e2990e |
hex | c47aeb67963 |
13502013405539 has 2 divisors, whose sum is σ = 13502013405540. Its totient is φ = 13502013405538.
The previous prime is 13502013405479. The next prime is 13502013405563. The reversal of 13502013405539 is 93550431020531.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13502013405539 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 13502013405496 and 13502013405505.
It is not a weakly prime, because it can be changed into another prime (13502013405589) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6751006702769 + 6751006702770.
It is an arithmetic number, because the mean of its divisors is an integer number (6751006702770).
Almost surely, 213502013405539 is an apocalyptic number.
13502013405539 is a deficient number, since it is larger than the sum of its proper divisors (1).
13502013405539 is an equidigital number, since it uses as much as digits as its factorization.
13502013405539 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 243000, while the sum is 41.
The spelling of 13502013405539 in words is "thirteen trillion, five hundred two billion, thirteen million, four hundred five thousand, five hundred thirty-nine".
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