Base | Representation |
---|---|
bin | 1110010011000010010101… |
… | …01010110011001001011011 |
3 | 11010022200012110111221200122 |
4 | 13021201022222303021123 |
5 | 13110104431303342001 |
6 | 150511305420023455 |
7 | 6423331515642536 |
oct | 711411252631133 |
9 | 133280173457618 |
10 | 31440413340251 |
11 | a021893753aa3 |
12 | 3639442991b8b |
13 | 1470a8392acb6 |
14 | 7a9a1cddba1d |
15 | 397c85d6bb1b |
hex | 1c984aab325b |
31440413340251 has 2 divisors, whose sum is σ = 31440413340252. Its totient is φ = 31440413340250.
The previous prime is 31440413340227. The next prime is 31440413340253. The reversal of 31440413340251 is 15204331404413.
It is a strong prime.
It is an emirp because it is prime and its reverse (15204331404413) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 31440413340251 - 26 = 31440413340187 is a prime.
Together with 31440413340253, 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 (31440413340253) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15720206670125 + 15720206670126.
It is an arithmetic number, because the mean of its divisors is an integer number (15720206670126).
Almost surely, 231440413340251 is an apocalyptic number.
31440413340251 is a deficient number, since it is larger than the sum of its proper divisors (1).
31440413340251 is an equidigital number, since it uses as much as digits as its factorization.
31440413340251 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 69120, while the sum is 35.
Adding to 31440413340251 its reverse (15204331404413), we get a palindrome (46644744744664).
The spelling of 31440413340251 in words is "thirty-one trillion, four hundred forty billion, four hundred thirteen million, three hundred forty thousand, two hundred fifty-one".
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