Base | Representation |
---|---|
bin | 101111111001101010… |
… | …010011000100100001 |
3 | 11220202110202002010211 |
4 | 233321222103010201 |
5 | 1320313403131301 |
6 | 35343401054121 |
7 | 3500364312226 |
oct | 577152230441 |
9 | 156673662124 |
10 | 51433255201 |
11 | 1a8a3805989 |
12 | 9b74a87341 |
13 | 4b08993023 |
14 | 26bcc1b94d |
15 | 1510608051 |
hex | bf9a93121 |
51433255201 has 4 divisors (see below), whose sum is σ = 53669483712. Its totient is φ = 49197026692.
The previous prime is 51433255171. The next prime is 51433255207. The reversal of 51433255201 is 10255233415.
51433255201 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a semiprime because it is the product of two primes, and also a Blum integer, because the two primes are equal to 3 mod 4.
It is a cyclic number.
It is not a de Polignac number, because 51433255201 - 29 = 51433254689 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (51433255207) by changing a digit.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 1118114221 + ... + 1118114266.
It is an arithmetic number, because the mean of its divisors is an integer number (13417370928).
Almost surely, 251433255201 is an apocalyptic number.
It is an amenable number.
51433255201 is a deficient number, since it is larger than the sum of its proper divisors (2236228511).
51433255201 is a wasteful number, since it uses less digits than its factorization.
51433255201 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 2236228510.
The product of its (nonzero) digits is 18000, while the sum is 31.
Adding to 51433255201 its reverse (10255233415), we get a palindrome (61688488616).
The spelling of 51433255201 in words is "fifty-one billion, four hundred thirty-three million, two hundred fifty-five thousand, two hundred one".
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