Base | Representation |
---|---|
bin | 1100010001111100000000… |
… | …0010001100101111111011 |
3 | 1202210210210210010020121012 |
4 | 3010133000002030233323 |
5 | 3232210204311411212 |
6 | 44414512310504135 |
7 | 2562336452306024 |
oct | 304370002145773 |
9 | 52723723106535 |
10 | 13502304013307 |
11 | 433632767a51a |
12 | 1620a0636504b |
13 | 76c353254c75 |
14 | 34972d23dc4b |
15 | 18635c6d0922 |
hex | c47c008cbfb |
13502304013307 has 2 divisors, whose sum is σ = 13502304013308. Its totient is φ = 13502304013306.
The previous prime is 13502304013291. The next prime is 13502304013339. The reversal of 13502304013307 is 70331040320531.
It is an a-pointer prime, because the next prime (13502304013339) can be obtained adding 13502304013307 to its sum of digits (32).
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13502304013307 - 24 = 13502304013291 is a prime.
It is not a weakly prime, because it can be changed into another prime (13502304013007) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6751152006653 + 6751152006654.
It is an arithmetic number, because the mean of its divisors is an integer number (6751152006654).
Almost surely, 213502304013307 is an apocalyptic number.
13502304013307 is a deficient number, since it is larger than the sum of its proper divisors (1).
13502304013307 is an equidigital number, since it uses as much as digits as its factorization.
13502304013307 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 22680, while the sum is 32.
Adding to 13502304013307 its reverse (70331040320531), we get a palindrome (83833344333838).
The spelling of 13502304013307 in words is "thirteen trillion, five hundred two billion, three hundred four million, thirteen thousand, three hundred seven".
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