Base | Representation |
---|---|
bin | 10101000111110101… |
… | …01010110010010011 |
3 | 1002021022012201002221 |
4 | 22203322222302103 |
5 | 141211020003103 |
6 | 5113131321511 |
7 | 551005261444 |
oct | 124372526223 |
9 | 32238181087 |
10 | 11340000403 |
11 | 489a153166 |
12 | 22458b0297 |
13 | 10b94bc214 |
14 | 7980d72cb |
15 | 4658501bd |
hex | 2a3eaac93 |
11340000403 has 2 divisors, whose sum is σ = 11340000404. Its totient is φ = 11340000402.
The previous prime is 11340000401. The next prime is 11340000407. The reversal of 11340000403 is 30400004311.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 11340000403 - 21 = 11340000401 is a prime.
Together with 11340000401, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (11340000401) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5670000201 + 5670000202.
It is an arithmetic number, because the mean of its divisors is an integer number (5670000202).
Almost surely, 211340000403 is an apocalyptic number.
11340000403 is a deficient number, since it is larger than the sum of its proper divisors (1).
11340000403 is an equidigital number, since it uses as much as digits as its factorization.
11340000403 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 144, while the sum is 16.
Adding to 11340000403 its reverse (30400004311), we get a palindrome (41740004714).
The spelling of 11340000403 in words is "eleven billion, three hundred forty million, four hundred three", and thus it is an aban number.
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