Base | Representation |
---|---|
bin | 1011101110010100111… |
… | …0100101101100011011 |
3 | 201020212220112200210011 |
4 | 2323211032211230123 |
5 | 11244444030134431 |
6 | 232310051302351 |
7 | 20360143215613 |
oct | 2734516455433 |
9 | 636786480704 |
10 | 201414302491 |
11 | 78468139636 |
12 | 330512aa9b7 |
13 | 15cbb33795c |
14 | 9a69c77a43 |
15 | 538c6e1cb1 |
hex | 2ee53a5b1b |
201414302491 has 2 divisors, whose sum is σ = 201414302492. Its totient is φ = 201414302490.
The previous prime is 201414302387. The next prime is 201414302543. The reversal of 201414302491 is 194203414102.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 201414302491 - 217 = 201414171419 is a prime.
It is not a weakly prime, because it can be changed into another prime (201414302291) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 100707151245 + 100707151246.
It is an arithmetic number, because the mean of its divisors is an integer number (100707151246).
Almost surely, 2201414302491 is an apocalyptic number.
201414302491 is a deficient number, since it is larger than the sum of its proper divisors (1).
201414302491 is an equidigital number, since it uses as much as digits as its factorization.
201414302491 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 6912, while the sum is 31.
Adding to 201414302491 its reverse (194203414102), we get a palindrome (395617716593).
The spelling of 201414302491 in words is "two hundred one billion, four hundred fourteen million, three hundred two thousand, four hundred ninety-one".
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