Base | Representation |
---|---|
bin | 1111011011110111001… |
… | …00001010010100000011 |
3 | 1212200221021201102221200 |
4 | 13231323210022110003 |
5 | 32142131114220103 |
6 | 1043350303552243 |
7 | 53213445036645 |
oct | 7557344122403 |
9 | 1780837642850 |
10 | 530354054403 |
11 | 194a16045878 |
12 | 869525bb683 |
13 | 3b020988604 |
14 | 1b9526b2895 |
15 | dbe08970a3 |
hex | 7b7b90a503 |
530354054403 has 6 divisors (see below), whose sum is σ = 766066967484. Its totient is φ = 353569369596.
The previous prime is 530354054383. The next prime is 530354054407. The reversal of 530354054403 is 304450453035.
530354054403 is a `hidden beast` number, since 530 + 35 + 4 + 0 + 54 + 40 + 3 = 666.
530354054403 is digitally balanced in base 4, because in such base it contains all the possibile digits an equal number of times.
It is a de Polignac number, because none of the positive numbers 2k-530354054403 is a prime.
It is not an unprimeable number, because it can be changed into a prime (530354054407) by changing a digit.
It is a polite number, since it can be written in 5 ways as a sum of consecutive naturals, for example, 29464114125 + ... + 29464114142.
It is an arithmetic number, because the mean of its divisors is an integer number (127677827914).
Almost surely, 2530354054403 is an apocalyptic number.
530354054403 is a deficient number, since it is larger than the sum of its proper divisors (235712913081).
530354054403 is a wasteful number, since it uses less digits than its factorization.
530354054403 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 58928228273 (or 58928228270 counting only the distinct ones).
The product of its (nonzero) digits is 216000, while the sum is 36.
The spelling of 530354054403 in words is "five hundred thirty billion, three hundred fifty-four million, fifty-four thousand, four hundred three".
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