Base | Representation |
---|---|
bin | 110100010001011111… |
… | …010101000110110001 |
3 | 12100212122201211212212 |
4 | 310101133111012301 |
5 | 1404422232003213 |
6 | 41441305501505 |
7 | 4024622063141 |
oct | 642137250661 |
9 | 170778654785 |
10 | 56128000433 |
11 | 2189287a51a |
12 | aa651a1895 |
13 | 53a650409b |
14 | 2a0653c921 |
15 | 16d78639a8 |
hex | d117d51b1 |
56128000433 has 2 divisors, whose sum is σ = 56128000434. Its totient is φ = 56128000432.
The previous prime is 56128000421. The next prime is 56128000463. The reversal of 56128000433 is 33400082165.
56128000433 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 29299853584 + 26828146849 = 171172^2 + 163793^2 .
It is a cyclic number.
It is not a de Polignac number, because 56128000433 - 24 = 56128000417 is a prime.
It is a self number, because there is not a number n which added to its sum of digits gives 56128000433.
It is not a weakly prime, because it can be changed into another prime (56128000463) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 28064000216 + 28064000217.
It is an arithmetic number, because the mean of its divisors is an integer number (28064000217).
Almost surely, 256128000433 is an apocalyptic number.
It is an amenable number.
56128000433 is a deficient number, since it is larger than the sum of its proper divisors (1).
56128000433 is an equidigital number, since it uses as much as digits as its factorization.
56128000433 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 17280, while the sum is 32.
Adding to 56128000433 its reverse (33400082165), we get a palindrome (89528082598).
The spelling of 56128000433 in words is "fifty-six billion, one hundred twenty-eight million, four hundred thirty-three", and thus it is an aban number.
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