Base | Representation |
---|---|
bin | 10010010100111000110011… |
… | …11001000101101111000111 |
3 | 12021200122122010121101022222 |
4 | 21022130121321011233013 |
5 | 20240234112302100341 |
6 | 221413341025242555 |
7 | 11326405233355046 |
oct | 1112343171055707 |
9 | 167618563541288 |
10 | 40300112534471 |
11 | 119281aa9a4169 |
12 | 462a52219345b |
13 | 19643908081ca |
14 | 9d477055255d |
15 | 49d47220944b |
hex | 24a719e45bc7 |
40300112534471 has 2 divisors, whose sum is σ = 40300112534472. Its totient is φ = 40300112534470.
The previous prime is 40300112534453. The next prime is 40300112534567. The reversal of 40300112534471 is 17443521100304.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 40300112534471 - 230 = 40299038792647 is a prime.
It is a super-3 number, since 3×403001125344713 (a number of 42 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (40300112534401) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 20150056267235 + 20150056267236.
It is an arithmetic number, because the mean of its divisors is an integer number (20150056267236).
Almost surely, 240300112534471 is an apocalyptic number.
40300112534471 is a deficient number, since it is larger than the sum of its proper divisors (1).
40300112534471 is an equidigital number, since it uses as much as digits as its factorization.
40300112534471 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 40320, while the sum is 35.
Adding to 40300112534471 its reverse (17443521100304), we get a palindrome (57743633634775).
The spelling of 40300112534471 in words is "forty trillion, three hundred billion, one hundred twelve million, five hundred thirty-four thousand, four hundred seventy-one".
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