Base | Representation |
---|---|
bin | 100110100110000110100… |
… | …0101011001101100000011 |
3 | 200210002212000100200101022 |
4 | 1031030031011121230003 |
5 | 1143402111024132212 |
6 | 15140504444500055 |
7 | 1055144362106654 |
oct | 115141505315403 |
9 | 20702760320338 |
10 | 5304504130307 |
11 | 176569a3255a8 |
12 | 71806b84a62b |
13 | 2c629c0b4625 |
14 | 144a4cb8b52b |
15 | 92eb0537772 |
hex | 4d30d159b03 |
5304504130307 has 2 divisors, whose sum is σ = 5304504130308. Its totient is φ = 5304504130306.
The previous prime is 5304504130201. The next prime is 5304504130313. The reversal of 5304504130307 is 7030314054035.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-5304504130307 is a prime.
It is a super-3 number, since 3×53045041303073 (a number of 39 digits) contains 333 as substring.
It is not a weakly prime, because it can be changed into another prime (5304504130397) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2652252065153 + 2652252065154.
It is an arithmetic number, because the mean of its divisors is an integer number (2652252065154).
Almost surely, 25304504130307 is an apocalyptic number.
5304504130307 is a deficient number, since it is larger than the sum of its proper divisors (1).
5304504130307 is an equidigital number, since it uses as much as digits as its factorization.
5304504130307 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 75600, while the sum is 35.
The spelling of 5304504130307 in words is "five trillion, three hundred four billion, five hundred four million, one hundred thirty thousand, three hundred seven".
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