Base | Representation |
---|---|
bin | 11001111011011010111011… |
… | …010011110001100111010011 |
3 | 112221202121001001200012011221 |
4 | 121323122323103301213103 |
5 | 104421320201200412042 |
6 | 1042310423123545511 |
7 | 33006502546616011 |
oct | 3173327323614723 |
9 | 487677031605157 |
10 | 114034524232147 |
11 | 33375852067422 |
12 | 109587a352b297 |
13 | 4b8254287130c |
14 | 202342c5018b1 |
15 | d2b4765bed67 |
hex | 67b6bb4f19d3 |
114034524232147 has 2 divisors, whose sum is σ = 114034524232148. Its totient is φ = 114034524232146.
The previous prime is 114034524232133. The next prime is 114034524232177. The reversal of 114034524232147 is 741232425430411.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 114034524232147 - 27 = 114034524232019 is a prime.
It is not a weakly prime, because it can be changed into another prime (114034524232127) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (29) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 57017262116073 + 57017262116074.
It is an arithmetic number, because the mean of its divisors is an integer number (57017262116074).
Almost surely, 2114034524232147 is an apocalyptic number.
114034524232147 is a deficient number, since it is larger than the sum of its proper divisors (1).
114034524232147 is an equidigital number, since it uses as much as digits as its factorization.
114034524232147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 645120, while the sum is 43.
Adding to 114034524232147 its reverse (741232425430411), we get a palindrome (855266949662558).
The spelling of 114034524232147 in words is "one hundred fourteen trillion, thirty-four billion, five hundred twenty-four million, two hundred thirty-two thousand, one hundred forty-seven".
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