Base | Representation |
---|---|
bin | 11000010011001001100000… |
… | …11001111100000000110001 |
3 | 21000012021210012021110020212 |
4 | 30021210300121330000301 |
5 | 24000432324030012223 |
6 | 305351245344513505 |
7 | 14153341504635521 |
oct | 1411446031740061 |
9 | 230167705243225 |
10 | 53434500235313 |
11 | 1603149839162a |
12 | 5babb6a01b895 |
13 | 23a7b0c3b2c22 |
14 | d2a35a679681 |
15 | 629e49a0e278 |
hex | 30993067c031 |
53434500235313 has 2 divisors, whose sum is σ = 53434500235314. Its totient is φ = 53434500235312.
The previous prime is 53434500235283. The next prime is 53434500235351. The reversal of 53434500235313 is 31353200543435.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 53434462572544 + 37662769 = 7309888^2 + 6137^2 .
It is a cyclic number.
It is not a de Polignac number, because 53434500235313 - 28 = 53434500235057 is a prime.
It is not a weakly prime, because it can be changed into another prime (53434500235373) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26717250117656 + 26717250117657.
It is an arithmetic number, because the mean of its divisors is an integer number (26717250117657).
Almost surely, 253434500235313 is an apocalyptic number.
It is an amenable number.
53434500235313 is a deficient number, since it is larger than the sum of its proper divisors (1).
53434500235313 is an equidigital number, since it uses as much as digits as its factorization.
53434500235313 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 972000, while the sum is 41.
Adding to 53434500235313 its reverse (31353200543435), we get a palindrome (84787700778748).
The spelling of 53434500235313 in words is "fifty-three trillion, four hundred thirty-four billion, five hundred million, two hundred thirty-five thousand, three hundred thirteen".
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