Base | Representation |
---|---|
bin | 101111100111011010… |
… | …100001111010010001 |
3 | 11212222010201101202211 |
4 | 233213122201322101 |
5 | 1314202021204133 |
6 | 35253143512121 |
7 | 3456656310163 |
oct | 574732417221 |
9 | 155863641684 |
10 | 51127131793 |
11 | 1a756a29931 |
12 | 9aaa458641 |
13 | 4a8a43000a |
14 | 26902d2933 |
15 | 14e37d9bcd |
hex | be76a1e91 |
51127131793 has 2 divisors, whose sum is σ = 51127131794. Its totient is φ = 51127131792.
The previous prime is 51127131719. The next prime is 51127131799. The reversal of 51127131793 is 39713172115.
It is a happy number.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 29703833104 + 21423298689 = 172348^2 + 146367^2 .
It is a cyclic number.
It is not a de Polignac number, because 51127131793 - 225 = 51093577361 is a prime.
It is a super-2 number, since 2×511271317932 (a number of 22 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (51127131799) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25563565896 + 25563565897.
It is an arithmetic number, because the mean of its divisors is an integer number (25563565897).
Almost surely, 251127131793 is an apocalyptic number.
It is an amenable number.
51127131793 is a deficient number, since it is larger than the sum of its proper divisors (1).
51127131793 is an equidigital number, since it uses as much as digits as its factorization.
51127131793 is an evil number, because the sum of its binary digits is even.
The product of its digits is 39690, while the sum is 40.
The spelling of 51127131793 in words is "fifty-one billion, one hundred twenty-seven million, one hundred thirty-one thousand, seven hundred ninety-three".
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