Base | Representation |
---|---|
bin | 11000001010000110010000… |
… | …00100001000111001001101 |
3 | 20222002120001220010112022002 |
4 | 30011003020010020321031 |
5 | 23430333423040421013 |
6 | 304552334324402045 |
7 | 14122022102635235 |
oct | 1405031004107115 |
9 | 228076056115262 |
10 | 53123512045133 |
11 | 15a21611693422 |
12 | 5b5b838846325 |
13 | 23846aac3324a |
14 | d1929805ccc5 |
15 | 621ce7845358 |
hex | 3050c8108e4d |
53123512045133 has 2 divisors, whose sum is σ = 53123512045134. Its totient is φ = 53123512045132.
The previous prime is 53123512045049. The next prime is 53123512045159. The reversal of 53123512045133 is 33154021532135.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 30932751605284 + 22190760439849 = 5561722^2 + 4710707^2 .
It is a cyclic number.
It is not a de Polignac number, because 53123512045133 - 212 = 53123512041037 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (53123512045183) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 26561756022566 + 26561756022567.
It is an arithmetic number, because the mean of its divisors is an integer number (26561756022567).
Almost surely, 253123512045133 is an apocalyptic number.
It is an amenable number.
53123512045133 is a deficient number, since it is larger than the sum of its proper divisors (1).
53123512045133 is an equidigital number, since it uses as much as digits as its factorization.
53123512045133 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 162000, while the sum is 38.
Adding to 53123512045133 its reverse (33154021532135), we get a palindrome (86277533577268).
The spelling of 53123512045133 in words is "fifty-three trillion, one hundred twenty-three billion, five hundred twelve million, forty-five thousand, one hundred thirty-three".
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