Base | Representation |
---|---|
bin | 1011110101010110111010… |
… | …1100010000000001011011 |
3 | 1201001212110202222011111002 |
4 | 2331111232230100001123 |
5 | 3201134132114031212 |
6 | 43401151550055215 |
7 | 2512015315655651 |
oct | 275255654200133 |
9 | 51055422864432 |
10 | 13011313033307 |
11 | 41670810040a6 |
12 | 156181a4aa50b |
13 | 734c666c4c74 |
14 | 32da726dcdd1 |
15 | 1786c295bac2 |
hex | bd56eb1005b |
13011313033307 has 2 divisors, whose sum is σ = 13011313033308. Its totient is φ = 13011313033306.
The previous prime is 13011313033289. The next prime is 13011313033421. The reversal of 13011313033307 is 70333031311031.
13011313033307 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13011313033307 - 228 = 13011044597851 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 13011313033307.
It is not a weakly prime, because it can be changed into another prime (13011313035307) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6505656516653 + 6505656516654.
It is an arithmetic number, because the mean of its divisors is an integer number (6505656516654).
Almost surely, 213011313033307 is an apocalyptic number.
13011313033307 is a deficient number, since it is larger than the sum of its proper divisors (1).
13011313033307 is an equidigital number, since it uses as much as digits as its factorization.
13011313033307 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5103, while the sum is 29.
Adding to 13011313033307 its reverse (70333031311031), we get a palindrome (83344344344338).
The spelling of 13011313033307 in words is "thirteen trillion, eleven billion, three hundred thirteen million, thirty-three thousand, three hundred seven".
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