Base | Representation |
---|---|
bin | 1100101111100011111100… |
… | …0110101000110110011011 |
3 | 1211121110110112202221222122 |
4 | 3023320333012220312123 |
5 | 3314030010424234342 |
6 | 45444401542411455 |
7 | 2644164432106541 |
oct | 313707706506633 |
9 | 54543415687878 |
10 | 14011242024347 |
11 | 45121529399a7 |
12 | 16a3580730b8b |
13 | 7a834005a969 |
14 | 36620d53dd91 |
15 | 1946e7d484d2 |
hex | cbe3f1a8d9b |
14011242024347 has 2 divisors, whose sum is σ = 14011242024348. Its totient is φ = 14011242024346.
The previous prime is 14011242024331. The next prime is 14011242024349. The reversal of 14011242024347 is 74342024211041.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 14011242024347 - 24 = 14011242024331 is a prime.
Together with 14011242024349, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (14011242024349) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7005621012173 + 7005621012174.
It is an arithmetic number, because the mean of its divisors is an integer number (7005621012174).
Almost surely, 214011242024347 is an apocalyptic number.
14011242024347 is a deficient number, since it is larger than the sum of its proper divisors (1).
14011242024347 is an equidigital number, since it uses as much as digits as its factorization.
14011242024347 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 43008, while the sum is 35.
Adding to 14011242024347 its reverse (74342024211041), we get a palindrome (88353266235388).
The spelling of 14011242024347 in words is "fourteen trillion, eleven billion, two hundred forty-two million, twenty-four thousand, three hundred forty-seven".
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