Base | Representation |
---|---|
bin | 111100010010101110111111… |
… | …1010101111110111100010011 |
3 | 2120112210001021112222021002220 |
4 | 1320211131333111332330103 |
5 | 1024003044121210241201 |
6 | 5115535003503112123 |
7 | 216464554405200555 |
oct | 17045357725767423 |
9 | 2515701245867086 |
10 | 530340403212051 |
11 | 143a8a113738363 |
12 | 4b593633ba2643 |
13 | 199bcb683ab6b3 |
14 | 94d68808b54d5 |
15 | 414a586e68036 |
hex | 1e2577f57ef13 |
530340403212051 has 8 divisors (see below), whose sum is σ = 708241172385024. Its totient is φ = 352999951423560.
The previous prime is 530340403212049. The next prime is 530340403212073. The reversal of 530340403212051 is 150212304043035.
It is a sphenic number, since it is the product of 3 distinct primes.
It is not a de Polignac number, because 530340403212051 - 21 = 530340403212049 is a prime.
It is not an unprimeable number, because it can be changed into a prime (530340403212751) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 140079343911 + ... + 140079347696.
It is an arithmetic number, because the mean of its divisors is an integer number (88530146548128).
Almost surely, 2530340403212051 is an apocalyptic number.
530340403212051 is a deficient number, since it is larger than the sum of its proper divisors (177900769172973).
530340403212051 is a wasteful number, since it uses less digits than its factorization.
530340403212051 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 280158692241.
The product of its (nonzero) digits is 43200, while the sum is 33.
Adding to 530340403212051 its reverse (150212304043035), we get a palindrome (680552707255086).
The spelling of 530340403212051 in words is "five hundred thirty trillion, three hundred forty billion, four hundred three million, two hundred twelve thousand, fifty-one".
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