Base | Representation |
---|---|
bin | 111100010010110100001010… |
… | …1011000111110010110010111 |
3 | 2120112211002221122022121122021 |
4 | 1320211220111120332112113 |
5 | 1024003234343142404142 |
6 | 5115544042003114011 |
7 | 216465426556534366 |
oct | 17045502530762627 |
9 | 2515732848277567 |
10 | 530351510513047 |
11 | 143a938a34a3a63 |
12 | 4b595813972907 |
13 | 199c0c18604a52 |
14 | 94d7215b177dd |
15 | 414a9d7150267 |
hex | 1e25a1563e597 |
530351510513047 has 2 divisors, whose sum is σ = 530351510513048. Its totient is φ = 530351510513046.
The previous prime is 530351510513003. The next prime is 530351510513053. The reversal of 530351510513047 is 740315015153035.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 530351510513047 - 219 = 530351509988759 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 530351510512994 and 530351510513012.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (530351510513347) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 265175755256523 + 265175755256524.
It is an arithmetic number, because the mean of its divisors is an integer number (265175755256524).
Almost surely, 2530351510513047 is an apocalyptic number.
530351510513047 is a deficient number, since it is larger than the sum of its proper divisors (1).
530351510513047 is an equidigital number, since it uses as much as digits as its factorization.
530351510513047 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 472500, while the sum is 43.
The spelling of 530351510513047 in words is "five hundred thirty trillion, three hundred fifty-one billion, five hundred ten million, five hundred thirteen thousand, forty-seven".
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