Base | Representation |
---|---|
bin | 111100011001110011000100… |
… | …1010111100010010101110011 |
3 | 2120200012212011102101012100212 |
4 | 1320303212021113202111303 |
5 | 1024120000401401001433 |
6 | 5122001002240134335 |
7 | 216624653451555014 |
oct | 17063461127422563 |
9 | 2520185142335325 |
10 | 531311234000243 |
11 | 144323913379734 |
12 | 4b70b8153073ab |
13 | 19a60585454998 |
14 | 952b85ad7500b |
15 | 41659579aed48 |
hex | 1e339895e2573 |
531311234000243 has 2 divisors, whose sum is σ = 531311234000244. Its totient is φ = 531311234000242.
The previous prime is 531311234000231. The next prime is 531311234000311. The reversal of 531311234000243 is 342000432113135.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 531311234000243 - 24 = 531311234000227 is a prime.
It is not a weakly prime, because it can be changed into another prime (531311234000203) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 265655617000121 + 265655617000122.
It is an arithmetic number, because the mean of its divisors is an integer number (265655617000122).
Almost surely, 2531311234000243 is an apocalyptic number.
531311234000243 is a deficient number, since it is larger than the sum of its proper divisors (1).
531311234000243 is an equidigital number, since it uses as much as digits as its factorization.
531311234000243 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 25920, while the sum is 32.
Adding to 531311234000243 its reverse (342000432113135), we get a palindrome (873311666113378).
The spelling of 531311234000243 in words is "five hundred thirty-one trillion, three hundred eleven billion, two hundred thirty-four million, two hundred forty-three", and thus it is an aban number.
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