Base | Representation |
---|---|
bin | 101000000110100010110… |
… | …0111111111001100011101 |
3 | 201111220102122202102122021 |
4 | 1100031011213333030131 |
5 | 1210300234423314013 |
6 | 15415555455020141 |
7 | 1106125601536213 |
oct | 120150547771435 |
9 | 21456378672567 |
10 | 5511611151133 |
11 | 183550889772a |
12 | 75022b50a651 |
13 | 30c98693aa69 |
14 | 150a98ac6cb3 |
15 | 985828b5a8d |
hex | 503459ff31d |
5511611151133 has 2 divisors, whose sum is σ = 5511611151134. Its totient is φ = 5511611151132.
The previous prime is 5511611151103. The next prime is 5511611151157. The reversal of 5511611151133 is 3311511161155.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 3929759099044 + 1581852052089 = 1982362^2 + 1257717^2 .
It is a cyclic number.
It is not a de Polignac number, because 5511611151133 - 217 = 5511611020061 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5511611151103) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2755805575566 + 2755805575567.
It is an arithmetic number, because the mean of its divisors is an integer number (2755805575567).
Almost surely, 25511611151133 is an apocalyptic number.
It is an amenable number.
5511611151133 is a deficient number, since it is larger than the sum of its proper divisors (1).
5511611151133 is an equidigital number, since it uses as much as digits as its factorization.
5511611151133 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 6750, while the sum is 34.
The spelling of 5511611151133 in words is "five trillion, five hundred eleven billion, six hundred eleven million, one hundred fifty-one thousand, one hundred thirty-three".
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