Base | Representation |
---|---|
bin | 111100010000011111100110… |
… | …1010000011000011111000011 |
3 | 2120111200121110100102120021212 |
4 | 1320200333031100120133003 |
5 | 1023433023001004040342 |
6 | 5115141320112040335 |
7 | 216433403532206111 |
oct | 17040771520303703 |
9 | 2514617410376255 |
10 | 530032472721347 |
11 | 143980566a2a012 |
12 | 4b543a168330ab |
13 | 19999b0356604b |
14 | 94c59cc400ab1 |
15 | 414256343e182 |
hex | 1e20fcd4187c3 |
530032472721347 has 2 divisors, whose sum is σ = 530032472721348. Its totient is φ = 530032472721346.
The previous prime is 530032472721299. The next prime is 530032472721373. The reversal of 530032472721347 is 743127274230035.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 530032472721347 - 210 = 530032472720323 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 530032472721295 and 530032472721304.
It is not a weakly prime, because it can be changed into another prime (530032472621347) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 265016236360673 + 265016236360674.
It is an arithmetic number, because the mean of its divisors is an integer number (265016236360674).
Almost surely, 2530032472721347 is an apocalyptic number.
530032472721347 is a deficient number, since it is larger than the sum of its proper divisors (1).
530032472721347 is an equidigital number, since it uses as much as digits as its factorization.
530032472721347 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5927040, while the sum is 50.
The spelling of 530032472721347 in words is "five hundred thirty trillion, thirty-two billion, four hundred seventy-two million, seven hundred twenty-one thousand, three hundred forty-seven".
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