Base | Representation |
---|---|
bin | 1100010011001100110… |
… | …0000100101010011011 |
3 | 202012102201012000002122 |
4 | 3010303030010222123 |
5 | 11430231410402342 |
6 | 241024145451455 |
7 | 21160341031643 |
oct | 3046314045233 |
9 | 665381160078 |
10 | 211312200347 |
11 | 81687256806 |
12 | 34b54040b8b |
13 | 16c07b26851 |
14 | a32861d523 |
15 | 576b6324d2 |
hex | 3133304a9b |
211312200347 has 2 divisors, whose sum is σ = 211312200348. Its totient is φ = 211312200346.
The previous prime is 211312200277. The next prime is 211312200349. The reversal of 211312200347 is 743002213112.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 211312200347 - 216 = 211312134811 is a prime.
It is a super-3 number, since 3×2113122003473 (a number of 35 digits) contains 333 as substring.
Together with 211312200349, it forms a pair of twin primes.
It is a Chen prime.
It is not a weakly prime, because it can be changed into another prime (211312200349) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (17) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 105656100173 + 105656100174.
It is an arithmetic number, because the mean of its divisors is an integer number (105656100174).
Almost surely, 2211312200347 is an apocalyptic number.
211312200347 is a deficient number, since it is larger than the sum of its proper divisors (1).
211312200347 is an equidigital number, since it uses as much as digits as its factorization.
211312200347 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 2016, while the sum is 26.
Adding to 211312200347 its reverse (743002213112), we get a palindrome (954314413459).
The spelling of 211312200347 in words is "two hundred eleven billion, three hundred twelve million, two hundred thousand, three hundred forty-seven".
• e-mail: info -at- numbersaplenty.com • Privacy notice • done in 0.070 sec. • engine limits •