Base | Representation |
---|---|
bin | 110011110110101011… |
… | …001110100010100001 |
3 | 12022201022020021202011 |
4 | 303312223032202201 |
5 | 1403012100300213 |
6 | 41324515352521 |
7 | 4010520222355 |
oct | 636653164241 |
9 | 168638207664 |
10 | 55678134433 |
11 | 21681945539 |
12 | a95a5b2741 |
13 | 533425339c |
14 | 29a28b7465 |
15 | 16ad10003d |
hex | cf6ace8a1 |
55678134433 has 2 divisors, whose sum is σ = 55678134434. Its totient is φ = 55678134432.
The previous prime is 55678134427. The next prime is 55678134497. The reversal of 55678134433 is 33443187655.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 54117647424 + 1560487009 = 232632^2 + 39503^2 .
It is a cyclic number.
It is not a de Polignac number, because 55678134433 - 213 = 55678126241 is a prime.
It is not a weakly prime, because it can be changed into another prime (55678534433) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 27839067216 + 27839067217.
It is an arithmetic number, because the mean of its divisors is an integer number (27839067217).
Almost surely, 255678134433 is an apocalyptic number.
It is an amenable number.
55678134433 is a deficient number, since it is larger than the sum of its proper divisors (1).
55678134433 is an equidigital number, since it uses as much as digits as its factorization.
55678134433 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 3628800, while the sum is 49.
The spelling of 55678134433 in words is "fifty-five billion, six hundred seventy-eight million, one hundred thirty-four thousand, four hundred thirty-three".
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