Base | Representation |
---|---|
bin | 11000111001001111… |
… | …00001010110000011 |
3 | 1021111102122120102202 |
4 | 30130213201112003 |
5 | 204332421233011 |
6 | 10050110503415 |
7 | 652124554634 |
oct | 143447412603 |
9 | 37442576382 |
10 | 13365024131 |
11 | 5739231498 |
12 | 270bb04b6b |
13 | 134cbb8504 |
14 | 90b02788b |
15 | 53350723b |
hex | 31c9e1583 |
13365024131 has 2 divisors, whose sum is σ = 13365024132. Its totient is φ = 13365024130.
The previous prime is 13365024103. The next prime is 13365024133. The reversal of 13365024131 is 13142056331.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 13365024131 - 26 = 13365024067 is a prime.
It is a super-2 number, since 2×133650241312 (a number of 21 digits) contains 22 as substring.
Together with 13365024133, it forms a pair of twin primes.
It is a Chen prime.
It is a junction number, because it is equal to n+sod(n) for n = 13365024094 and 13365024103.
It is not a weakly prime, because it can be changed into another prime (13365024133) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6682512065 + 6682512066.
It is an arithmetic number, because the mean of its divisors is an integer number (6682512066).
Almost surely, 213365024131 is an apocalyptic number.
13365024131 is a deficient number, since it is larger than the sum of its proper divisors (1).
13365024131 is an equidigital number, since it uses as much as digits as its factorization.
13365024131 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6480, while the sum is 29.
The spelling of 13365024131 in words is "thirteen billion, three hundred sixty-five million, twenty-four thousand, one hundred thirty-one".
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