Base | Representation |
---|---|
bin | 100101110001011011… |
… | …0101010001100100001 |
3 | 21202101001220210110121 |
4 | 1023202312222030201 |
5 | 2312111020001003 |
6 | 101133000200241 |
7 | 5601054305455 |
oct | 1134266521441 |
9 | 252331823417 |
10 | 81115390753 |
11 | 31445583721 |
12 | 13879469081 |
13 | 7859243c47 |
14 | 3cd6d62665 |
15 | 219b394cbd |
hex | 12e2daa321 |
81115390753 has 2 divisors, whose sum is σ = 81115390754. Its totient is φ = 81115390752.
The previous prime is 81115390747. The next prime is 81115390777. The reversal of 81115390753 is 35709351118.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 78230411809 + 2884978944 = 279697^2 + 53712^2 .
It is a cyclic number.
It is not a de Polignac number, because 81115390753 - 29 = 81115390241 is a prime.
It is a super-3 number, since 3×811153907533 (a number of 34 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (81115390723) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 40557695376 + 40557695377.
It is an arithmetic number, because the mean of its divisors is an integer number (40557695377).
Almost surely, 281115390753 is an apocalyptic number.
It is an amenable number.
81115390753 is a deficient number, since it is larger than the sum of its proper divisors (1).
81115390753 is an equidigital number, since it uses as much as digits as its factorization.
81115390753 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 113400, while the sum is 43.
The spelling of 81115390753 in words is "eighty-one billion, one hundred fifteen million, three hundred ninety thousand, seven hundred fifty-three".
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