Base | Representation |
---|---|
bin | 111100010010110100001010… |
… | …1011000111110010110011011 |
3 | 2120112211002221122022121122102 |
4 | 1320211220111120332112123 |
5 | 1024003234343142404201 |
6 | 5115544042003114015 |
7 | 216465426556534403 |
oct | 17045502530762633 |
9 | 2515732848277572 |
10 | 530351510513051 |
11 | 143a938a34a3a67 |
12 | 4b59581397290b |
13 | 199c0c18604a56 |
14 | 94d7215b17803 |
15 | 414a9d715026b |
hex | 1e25a1563e59b |
530351510513051 has 16 divisors (see below), whose sum is σ = 545543904288768. Its totient is φ = 515368815559200.
The previous prime is 530351510513047. The next prime is 530351510513053. The reversal of 530351510513051 is 150315015153035.
It is a cyclic number.
It is not a de Polignac number, because 530351510513051 - 22 = 530351510513047 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 530351510512996 and 530351510513014.
It is not an unprimeable number, because it can be changed into a prime (530351510513053) by changing a digit.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 1155202790 + ... + 1155661796.
It is an arithmetic number, because the mean of its divisors is an integer number (34096494018048).
Almost surely, 2530351510513051 is an apocalyptic number.
530351510513051 is a deficient number, since it is larger than the sum of its proper divisors (15192393775717).
530351510513051 is a wasteful number, since it uses less digits than its factorization.
530351510513051 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 687362.
The product of its (nonzero) digits is 84375, while the sum is 38.
Adding to 530351510513051 its reverse (150315015153035), we get a palindrome (680666525666086).
The spelling of 530351510513051 in words is "five hundred thirty trillion, three hundred fifty-one billion, five hundred ten million, five hundred thirteen thousand, fifty-one".
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