Base | Representation |
---|---|
bin | 1110010010110100100011… |
… | …00110011011011101110111 |
3 | 11010021222002202022012012201 |
4 | 13021122101212123131313 |
5 | 13104444302420124143 |
6 | 150504043234325331 |
7 | 6422650242565435 |
oct | 711322146333567 |
9 | 133258082265181 |
10 | 31433013442423 |
11 | a0197366a425a |
12 | 3637b18729847 |
13 | 147017481b128 |
14 | 7a951a305d55 |
15 | 3979a13ae64d |
hex | 1c969199b777 |
31433013442423 has 2 divisors, whose sum is σ = 31433013442424. Its totient is φ = 31433013442422.
The previous prime is 31433013442417. The next prime is 31433013442463. The reversal of 31433013442423 is 32424431033413.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 31433013442423 - 213 = 31433013434231 is a prime.
It is a super-3 number, since 3×314330134424233 (a number of 41 digits) contains 333 as substring. Note that it is a super-d number also for d = 2.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (31433013442463) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 15716506721211 + 15716506721212.
It is an arithmetic number, because the mean of its divisors is an integer number (15716506721212).
Almost surely, 231433013442423 is an apocalyptic number.
31433013442423 is a deficient number, since it is larger than the sum of its proper divisors (1).
31433013442423 is an equidigital number, since it uses as much as digits as its factorization.
31433013442423 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 248832, while the sum is 37.
Adding to 31433013442423 its reverse (32424431033413), we get a palindrome (63857444475836).
The spelling of 31433013442423 in words is "thirty-one trillion, four hundred thirty-three billion, thirteen million, four hundred forty-two thousand, four hundred twenty-three".
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