Base | Representation |
---|---|
bin | 1111011011101011010… |
… | …00001010100010111011 |
3 | 1212200200100021112112212 |
4 | 13231311220022202323 |
5 | 32141430114100020 |
6 | 1043332344525335 |
7 | 53211125243543 |
oct | 7556550124273 |
9 | 1780610245485 |
10 | 530254440635 |
11 | 194974898547 |
12 | 8692518084b |
13 | 3b00614b85a |
14 | 1b943382323 |
15 | dbd6c6bcc5 |
hex | 7b75a0a8bb |
530254440635 has 4 divisors (see below), whose sum is σ = 636305328768. Its totient is φ = 424203552504.
The previous prime is 530254440619. The next prime is 530254440647. The reversal of 530254440635 is 536044452035.
It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 536044452035 = 5 ⋅107208890407.
It is a cyclic number.
It is not a de Polignac number, because 530254440635 - 24 = 530254440619 is a prime.
It is a Duffinian number.
It is a junction number, because it is equal to n+sod(n) for n = 530254440592 and 530254440601.
It is an unprimeable number.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 53025444059 + ... + 53025444068.
It is an arithmetic number, because the mean of its divisors is an integer number (159076332192).
Almost surely, 2530254440635 is an apocalyptic number.
530254440635 is a deficient number, since it is larger than the sum of its proper divisors (106050888133).
530254440635 is a wasteful number, since it uses less digits than its factorization.
530254440635 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 106050888132.
The product of its (nonzero) digits is 864000, while the sum is 41.
The spelling of 530254440635 in words is "five hundred thirty billion, two hundred fifty-four million, four hundred forty thousand, six hundred thirty-five".
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