Base | Representation |
---|---|
bin | 1111011011110100100… |
… | …01100011100110101001 |
3 | 1212200212202222021202021 |
4 | 13231322101203212221 |
5 | 32142110043300131 |
6 | 1043344211504441 |
7 | 53213063010463 |
oct | 7557221434651 |
9 | 1780782867667 |
10 | 530332400041 |
11 | 194a048a5604 |
12 | 869472b8121 |
13 | 3b019346192 |
14 | 1b94d877133 |
15 | dbdda1ae11 |
hex | 7b7a4639a9 |
530332400041 has 2 divisors, whose sum is σ = 530332400042. Its totient is φ = 530332400040.
The previous prime is 530332400029. The next prime is 530332400057. The reversal of 530332400041 is 140004233035.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 480104866816 + 50227533225 = 692896^2 + 224115^2 .
It is a cyclic number.
It is not a de Polignac number, because 530332400041 - 215 = 530332367273 is a prime.
It is a junction number, because it is equal to n+sod(n) for n = 530332399979 and 530332400015.
It is not a weakly prime, because it can be changed into another prime (530332400021) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 265166200020 + 265166200021.
It is an arithmetic number, because the mean of its divisors is an integer number (265166200021).
Almost surely, 2530332400041 is an apocalyptic number.
It is an amenable number.
530332400041 is a deficient number, since it is larger than the sum of its proper divisors (1).
530332400041 is an equidigital number, since it uses as much as digits as its factorization.
530332400041 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 4320, while the sum is 25.
Adding to 530332400041 its reverse (140004233035), we get a palindrome (670336633076).
The spelling of 530332400041 in words is "five hundred thirty billion, three hundred thirty-two million, four hundred thousand, forty-one".
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