Base | Representation |
---|---|
bin | 100101001011101110011… |
… | …1100101001010001110111 |
3 | 200002112220011110201221201 |
4 | 1022113130330221101313 |
5 | 1132212121242001213 |
6 | 14511410152353331 |
7 | 1035134021530234 |
oct | 112273474512167 |
9 | 20075804421851 |
10 | 5110423000183 |
11 | 16a03556119a6 |
12 | 6a65263a5847 |
13 | 2b0bac0c6627 |
14 | 1394babda78b |
15 | 8ce01a226dd |
hex | 4a5dcf29477 |
5110423000183 has 2 divisors, whose sum is σ = 5110423000184. Its totient is φ = 5110423000182.
The previous prime is 5110423000139. The next prime is 5110423000187. The reversal of 5110423000183 is 3810003240115.
It is a happy number.
5110423000183 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 5110423000183 - 221 = 5110420903031 is a prime.
It is a super-2 number, since 2×51104230001832 (a number of 26 digits) contains 22 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5110423000187) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2555211500091 + 2555211500092.
It is an arithmetic number, because the mean of its divisors is an integer number (2555211500092).
Almost surely, 25110423000183 is an apocalyptic number.
5110423000183 is a deficient number, since it is larger than the sum of its proper divisors (1).
5110423000183 is an equidigital number, since it uses as much as digits as its factorization.
5110423000183 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 2880, while the sum is 28.
Adding to 5110423000183 its reverse (3810003240115), we get a palindrome (8920426240298).
The spelling of 5110423000183 in words is "five trillion, one hundred ten billion, four hundred twenty-three million, one hundred eighty-three", and thus it is an aban number.
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