Base | Representation |
---|---|
bin | 11000101011101100… |
… | …00011101100100011 |
3 | 1021012111211120122221 |
4 | 30111312003230203 |
5 | 204114324104324 |
6 | 10030531211511 |
7 | 646244664511 |
oct | 142566035443 |
9 | 37174746587 |
10 | 13251394339 |
11 | 5690080777 |
12 | 2699a46b97 |
13 | 13324b1c51 |
14 | 8d9cc94b1 |
15 | 52855e0e4 |
hex | 315d83b23 |
13251394339 has 4 divisors (see below), whose sum is σ = 13310817856. Its totient is φ = 13191970824.
The previous prime is 13251394337. The next prime is 13251394369. The reversal of 13251394339 is 93349315231.
13251394339 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a semiprime because it is the product of two primes.
It is a cyclic number.
It is not a de Polignac number, because 13251394339 - 21 = 13251394337 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 13251394295 and 13251394304.
It is not an unprimeable number, because it can be changed into a prime (13251394337) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 29711424 + ... + 29711869.
It is an arithmetic number, because the mean of its divisors is an integer number (3327704464).
Almost surely, 213251394339 is an apocalyptic number.
13251394339 is a deficient number, since it is larger than the sum of its proper divisors (59423517).
13251394339 is an equidigital number, since it uses as much as digits as its factorization.
13251394339 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 59423516.
The product of its digits is 262440, while the sum is 43.
The spelling of 13251394339 in words is "thirteen billion, two hundred fifty-one million, three hundred ninety-four thousand, three hundred thirty-nine".
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