Base | Representation |
---|---|
bin | 1111011011110111001… |
… | …00000000010010100001 |
3 | 1212200221021122100122000 |
4 | 13231323210000102201 |
5 | 32142131111411340 |
6 | 1043350303034213 |
7 | 53213444506152 |
oct | 7557344002241 |
9 | 1780837570560 |
10 | 530354013345 |
11 | 194a16017a42 |
12 | 8695259b969 |
13 | 3b020972a10 |
14 | 1b9526a1929 |
15 | dbe0889d30 |
hex | 7b7b9004a1 |
530354013345 has 64 divisors (see below), whose sum is σ = 1021459299840. Its totient is φ = 259533759744.
The previous prime is 530354013281. The next prime is 530354013347. The reversal of 530354013345 is 543310453035.
530354013345 is a `hidden beast` number, since 5 + 303 + 5 + 4 + 0 + 1 + 3 + 345 = 666.
It is not a de Polignac number, because 530354013345 - 26 = 530354013281 is a prime.
It is a super-3 number, since 3×5303540133453 (a number of 36 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is not an unprimeable number, because it can be changed into a prime (530354013347) by changing a digit.
It is a polite number, since it can be written in 63 ways as a sum of consecutive naturals, for example, 611694 + ... + 1197863.
It is an arithmetic number, because the mean of its divisors is an integer number (15960301560).
Almost surely, 2530354013345 is an apocalyptic number.
It is an amenable number.
530354013345 is a deficient number, since it is larger than the sum of its proper divisors (491105286495).
530354013345 is a wasteful number, since it uses less digits than its factorization.
530354013345 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1809751 (or 1809745 counting only the distinct ones).
The product of its (nonzero) digits is 162000, while the sum is 36.
The spelling of 530354013345 in words is "five hundred thirty billion, three hundred fifty-four million, thirteen thousand, three hundred forty-five".
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