Base | Representation |
---|---|
bin | 1011111010110101011100… |
… | …0010101001000011011011 |
3 | 1201101212100020121021010021 |
4 | 2332231113002221003123 |
5 | 3204204332303432311 |
6 | 43512312402531311 |
7 | 2521556111001502 |
oct | 276552702510333 |
9 | 51355306537107 |
10 | 13105405530331 |
11 | 41a2a75845834 |
12 | 1577ab9966b37 |
13 | 740ab03532ca |
14 | 33443ad4b839 |
15 | 17ad7d230171 |
hex | beb570a90db |
13105405530331 has 2 divisors, whose sum is σ = 13105405530332. Its totient is φ = 13105405530330.
The previous prime is 13105405530323. The next prime is 13105405530337. The reversal of 13105405530331 is 13303550450131.
It is a happy number.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13105405530331 - 23 = 13105405530323 is a prime.
It is a super-2 number, since 2×131054055303312 (a number of 27 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (13105405530337) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6552702765165 + 6552702765166.
It is an arithmetic number, because the mean of its divisors is an integer number (6552702765166).
Almost surely, 213105405530331 is an apocalyptic number.
13105405530331 is a deficient number, since it is larger than the sum of its proper divisors (1).
13105405530331 is an equidigital number, since it uses as much as digits as its factorization.
13105405530331 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 40500, while the sum is 34.
Adding to 13105405530331 its reverse (13303550450131), we get a palindrome (26408955980462).
The spelling of 13105405530331 in words is "thirteen trillion, one hundred five billion, four hundred five million, five hundred thirty thousand, three hundred thirty-one".
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