Base | Representation |
---|---|
bin | 100011011001000110001001… |
… | …0011101011001101010010111 |
3 | 1111211021011210002212201021222 |
4 | 1012302030102131121222113 |
5 | 311301020321044310033 |
6 | 3022034541531000555 |
7 | 122400403040651603 |
oct | 10662142235315227 |
9 | 1454234702781258 |
10 | 311312424213143 |
11 | 9021494862a981 |
12 | 2aaba524b9915b |
13 | 1049283565a882 |
14 | 56c385d1d6503 |
15 | 25ece3d2e0598 |
hex | 11b2312759a97 |
311312424213143 has 2 divisors, whose sum is σ = 311312424213144. Its totient is φ = 311312424213142.
The previous prime is 311312424213067. The next prime is 311312424213151. The reversal of 311312424213143 is 341312424213113.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 311312424213143 - 214 = 311312424196759 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (341312424213143) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 155656212106571 + 155656212106572.
It is an arithmetic number, because the mean of its divisors is an integer number (155656212106572).
Almost surely, 2311312424213143 is an apocalyptic number.
311312424213143 is a deficient number, since it is larger than the sum of its proper divisors (1).
311312424213143 is an equidigital number, since it uses as much as digits as its factorization.
311312424213143 is an evil number, because the sum of its binary digits is even.
The product of its digits is 41472, while the sum is 35.
Adding to 311312424213143 its reverse (341312424213113), we get a palindrome (652624848426256).
The spelling of 311312424213143 in words is "three hundred eleven trillion, three hundred twelve billion, four hundred twenty-four million, two hundred thirteen thousand, one hundred forty-three".
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