Base | Representation |
---|---|
bin | 1100000111111011010100… |
… | …1011001000000011010111 |
3 | 1202012100220010201020221102 |
4 | 3001332311023020003113 |
5 | 3221401000204042101 |
6 | 44203511011140315 |
7 | 2544040555622603 |
oct | 301766513100327 |
9 | 52170803636842 |
10 | 13330323112151 |
11 | 427a3a396aa61 |
12 | 15b360a2a469b |
13 | 759075aaca26 |
14 | 341296480503 |
15 | 181b43ee686b |
hex | c1fb52c80d7 |
13330323112151 has 2 divisors, whose sum is σ = 13330323112152. Its totient is φ = 13330323112150.
The previous prime is 13330323112133. The next prime is 13330323112159. The reversal of 13330323112151 is 15121132303331.
13330323112151 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-13330323112151 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (13330323112159) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6665161556075 + 6665161556076.
It is an arithmetic number, because the mean of its divisors is an integer number (6665161556076).
Almost surely, 213330323112151 is an apocalyptic number.
13330323112151 is a deficient number, since it is larger than the sum of its proper divisors (1).
13330323112151 is an equidigital number, since it uses as much as digits as its factorization.
13330323112151 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 4860, while the sum is 29.
Adding to 13330323112151 its reverse (15121132303331), we get a palindrome (28451455415482).
The spelling of 13330323112151 in words is "thirteen trillion, three hundred thirty billion, three hundred twenty-three million, one hundred twelve thousand, one hundred fifty-one".
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