Base | Representation |
---|---|
bin | 100110100101110111111… |
… | …0001000010111101011111 |
3 | 200210001120201210211101222 |
4 | 1031023233301002331133 |
5 | 1143400104114130241 |
6 | 15140343554123555 |
7 | 1055126240202245 |
oct | 115135761027537 |
9 | 20701521724358 |
10 | 5304012255071 |
11 | 1765467708442 |
12 | 717b52b805bb |
13 | 2c62212156aa |
14 | 144a0370c795 |
15 | 92e82276a4b |
hex | 4d2efc42f5f |
5304012255071 has 2 divisors, whose sum is σ = 5304012255072. Its totient is φ = 5304012255070.
The previous prime is 5304012255043. The next prime is 5304012255107. The reversal of 5304012255071 is 1705522104035.
5304012255071 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
Together with next prime (5304012255107) it forms an Ormiston pair, because they use the same digits, order apart.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 5304012255071 - 226 = 5303945146207 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (5304012251071) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2652006127535 + 2652006127536.
It is an arithmetic number, because the mean of its divisors is an integer number (2652006127536).
Almost surely, 25304012255071 is an apocalyptic number.
5304012255071 is a deficient number, since it is larger than the sum of its proper divisors (1).
5304012255071 is an equidigital number, since it uses as much as digits as its factorization.
5304012255071 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 42000, while the sum is 35.
The spelling of 5304012255071 in words is "five trillion, three hundred four billion, twelve million, two hundred fifty-five thousand, seventy-one".
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