Base | Representation |
---|---|
bin | 101111100010011110… |
… | …100110100001001001 |
3 | 11212202100212202222002 |
4 | 233202132212201021 |
5 | 1314014321010121 |
6 | 35241024440345 |
7 | 3454632334352 |
oct | 574236464111 |
9 | 155670782862 |
10 | 51044313161 |
11 | 1a714202a86 |
12 | 9a867790b5 |
13 | 4a762239ca |
14 | 26832d4d29 |
15 | 14db3cae0b |
hex | be27a6849 |
51044313161 has 2 divisors, whose sum is σ = 51044313162. Its totient is φ = 51044313160.
The previous prime is 51044313113. The next prime is 51044313217. The reversal of 51044313161 is 16131344015.
51044313161 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 45866647225 + 5177665936 = 214165^2 + 71956^2 .
It is a cyclic number.
It is not a de Polignac number, because 51044313161 - 26 = 51044313097 is a prime.
It is not a weakly prime, because it can be changed into another prime (51044213161) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 25522156580 + 25522156581.
It is an arithmetic number, because the mean of its divisors is an integer number (25522156581).
Almost surely, 251044313161 is an apocalyptic number.
It is an amenable number.
51044313161 is a deficient number, since it is larger than the sum of its proper divisors (1).
51044313161 is an equidigital number, since it uses as much as digits as its factorization.
51044313161 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 4320, while the sum is 29.
Adding to 51044313161 its reverse (16131344015), we get a palindrome (67175657176).
The spelling of 51044313161 in words is "fifty-one billion, forty-four million, three hundred thirteen thousand, one hundred sixty-one".
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