Base | Representation |
---|---|
bin | 110001011111011001… |
… | …000110000100011101 |
3 | 12002011102022210101200 |
4 | 301133121012010131 |
5 | 1332312321142113 |
6 | 40225011030113 |
7 | 3560600505324 |
oct | 613731060435 |
9 | 162142283350 |
10 | 53140021533 |
11 | 2059a17a775 |
12 | a37059b339 |
13 | 501b46b331 |
14 | 28017850bb |
15 | 15b0396873 |
hex | c5f64611d |
53140021533 has 24 divisors (see below), whose sum is σ = 79449916416. Its totient is φ = 34190402400.
The previous prime is 53140021531. The next prime is 53140021547. The reversal of 53140021533 is 33512004135.
53140021533 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is not a de Polignac number, because 53140021533 - 21 = 53140021531 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 53140021497 and 53140021506.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (53140021531) by changing a digit.
It is a polite number, since it can be written in 23 ways as a sum of consecutive naturals, for example, 157098 + ... + 361883.
It is an arithmetic number, because the mean of its divisors is an integer number (3310413184).
Almost surely, 253140021533 is an apocalyptic number.
It is an amenable number.
53140021533 is a deficient number, since it is larger than the sum of its proper divisors (26309894883).
53140021533 is a wasteful number, since it uses less digits than its factorization.
53140021533 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 519385 (or 519382 counting only the distinct ones).
The product of its (nonzero) digits is 5400, while the sum is 27.
Adding to 53140021533 its reverse (33512004135), we get a palindrome (86652025668).
The spelling of 53140021533 in words is "fifty-three billion, one hundred forty million, twenty-one thousand, five hundred thirty-three".
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