Base | Representation |
---|---|
bin | 1001101001011010001111… |
… | …11011110110010111100111 |
3 | 2210010001011122101011120211 |
4 | 10310231013323312113213 |
5 | 10240032333342041341 |
6 | 113041333350240251 |
7 | 4316443641503113 |
oct | 464550773662747 |
9 | 83101148334524 |
10 | 21214050346471 |
11 | 683991240a414 |
12 | 2467512656087 |
13 | bab626757608 |
14 | 534aa0806743 |
15 | 26bc5d2ec181 |
hex | 134b47ef65e7 |
21214050346471 has 2 divisors, whose sum is σ = 21214050346472. Its totient is φ = 21214050346470.
The previous prime is 21214050346421. The next prime is 21214050346529. The reversal of 21214050346471 is 17464305041212.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 21214050346471 - 211 = 21214050344423 is a prime.
It is a super-3 number, since 3×212140503464713 (a number of 41 digits) contains 333 as substring.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (21214050346421) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 10607025173235 + 10607025173236.
It is an arithmetic number, because the mean of its divisors is an integer number (10607025173236).
Almost surely, 221214050346471 is an apocalyptic number.
21214050346471 is a deficient number, since it is larger than the sum of its proper divisors (1).
21214050346471 is an equidigital number, since it uses as much as digits as its factorization.
21214050346471 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 161280, while the sum is 40.
Adding to 21214050346471 its reverse (17464305041212), we get a palindrome (38678355387683).
The spelling of 21214050346471 in words is "twenty-one trillion, two hundred fourteen billion, fifty million, three hundred forty-six thousand, four hundred seventy-one".
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