Base | Representation |
---|---|
bin | 1101000001110001111011… |
… | …0011001001010100000011 |
3 | 1212201101100101212110102021 |
4 | 3100130132303021110003 |
5 | 3334142012024444011 |
6 | 50244243414102311 |
7 | 3005614560431236 |
oct | 320343663112403 |
9 | 55641311773367 |
10 | 14324232656131 |
11 | 46229669a2122 |
12 | 173417064a997 |
13 | 7cba0032b128 |
14 | 377421a8d41d |
15 | 19c915c08471 |
hex | d071ecc9503 |
14324232656131 has 2 divisors, whose sum is σ = 14324232656132. Its totient is φ = 14324232656130.
The previous prime is 14324232656033. The next prime is 14324232656207. The reversal of 14324232656131 is 13165623242341.
It is a strong prime.
It is an emirp because it is prime and its reverse (13165623242341) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 14324232656131 - 243 = 5528139633923 is a prime.
It is a super-2 number, since 2×143242326561312 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (14324232686131) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7162116328065 + 7162116328066.
It is an arithmetic number, because the mean of its divisors is an integer number (7162116328066).
Almost surely, 214324232656131 is an apocalyptic number.
14324232656131 is a deficient number, since it is larger than the sum of its proper divisors (1).
14324232656131 is an equidigital number, since it uses as much as digits as its factorization.
14324232656131 is an evil number, because the sum of its binary digits is even.
The product of its digits is 622080, while the sum is 43.
Adding to 14324232656131 its reverse (13165623242341), we get a palindrome (27489855898472).
The spelling of 14324232656131 in words is "fourteen trillion, three hundred twenty-four billion, two hundred thirty-two million, six hundred fifty-six thousand, one hundred thirty-one".
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