Base | Representation |
---|---|
bin | 111100010010101111110010… |
… | …1000100111111100011100110 |
3 | 2120112210012200111122210212012 |
4 | 1320211133211010333203212 |
5 | 1024003111120132043100 |
6 | 5115535445114154222 |
7 | 216464644612015361 |
oct | 17045374504774346 |
9 | 2515705614583765 |
10 | 530342110034150 |
11 | 143a8a90a140037 |
12 | 4b593a2b722972 |
13 | 199c007aba61bb |
14 | 94d69a34550d8 |
15 | 414a636c18135 |
hex | 1e257e513f8e6 |
530342110034150 has 72 divisors (see below), whose sum is σ = 990173201728872. Its totient is φ = 211335802128000.
The previous prime is 530342110034137. The next prime is 530342110034179. The reversal of 530342110034150 is 51430011243035.
It is a super-2 number, since 2×5303421100341502 (a number of 30 digits) contains 22 as substring.
It is a Curzon number.
It is a congruent number.
It is an unprimeable number.
It is a polite number, since it can be written in 35 ways as a sum of consecutive naturals, for example, 6418001234 + ... + 6418083866.
Almost surely, 2530342110034150 is an apocalyptic number.
530342110034150 is a gapful number since it is divisible by the number (50) formed by its first and last digit.
530342110034150 is a deficient number, since it is larger than the sum of its proper divisors (459831091694722).
530342110034150 is an equidigital number, since it uses as much as digits as its factorization.
530342110034150 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 84198 (or 83762 counting only the distinct ones).
The product of its (nonzero) digits is 21600, while the sum is 32.
Adding to 530342110034150 its reverse (51430011243035), we get a palindrome (581772121277185).
The spelling of 530342110034150 in words is "five hundred thirty trillion, three hundred forty-two billion, one hundred ten million, thirty-four thousand, one hundred fifty".
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