Base | Representation |
---|---|
bin | 10011000001100001… |
… | …10011010101010001 |
3 | 222100210012110021021 |
4 | 21200300303111101 |
5 | 131404103343442 |
6 | 4405243155441 |
7 | 511044650623 |
oct | 114060632521 |
9 | 28323173237 |
10 | 10213340497 |
11 | 4371189978 |
12 | 1b90515b81 |
13 | c69c65959 |
14 | 6cc618c13 |
15 | 3eb99ee67 |
hex | 260c33551 |
10213340497 has 2 divisors, whose sum is σ = 10213340498. Its totient is φ = 10213340496.
The previous prime is 10213340479. The next prime is 10213340513. The reversal of 10213340497 is 79404331201.
10213340497 is digitally balanced in base 3, because in such base it contains all the possibile digits an equal number of times.
Together with previous prime (10213340479) it forms an Ormiston pair, because they use the same digits, order apart.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 8559210256 + 1654130241 = 92516^2 + 40671^2 .
It is a cyclic number.
It is not a de Polignac number, because 10213340497 - 211 = 10213338449 is a prime.
It is not a weakly prime, because it can be changed into another prime (10213340447) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5106670248 + 5106670249.
It is an arithmetic number, because the mean of its divisors is an integer number (5106670249).
Almost surely, 210213340497 is an apocalyptic number.
It is an amenable number.
10213340497 is a deficient number, since it is larger than the sum of its proper divisors (1).
10213340497 is an equidigital number, since it uses as much as digits as its factorization.
10213340497 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 18144, while the sum is 34.
Adding to 10213340497 its reverse (79404331201), we get a palindrome (89617671698).
The spelling of 10213340497 in words is "ten billion, two hundred thirteen million, three hundred forty thousand, four hundred ninety-seven".
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