Base | Representation |
---|---|
bin | 110000000001100001101010… |
… | …101101101111000010100011 |
3 | 1000200211120202212011122212101 |
4 | 300001201222231233002203 |
5 | 210140440314120402042 |
6 | 2025113012452111231 |
7 | 62326333002112201 |
oct | 6001415255570243 |
9 | 1020746685148771 |
10 | 211211102122147 |
11 | 61331141982116 |
12 | 1b832162905517 |
13 | 90b1176b3413a |
14 | 3a22959932071 |
15 | 19641436d06b7 |
hex | c0186ab6f0a3 |
211211102122147 has 2 divisors, whose sum is σ = 211211102122148. Its totient is φ = 211211102122146.
The previous prime is 211211102122091. The next prime is 211211102122187. The reversal of 211211102122147 is 741221201112112.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 211211102122147 - 235 = 211176742383779 is a prime.
It is a super-2 number, since 2×2112111021221472 (a number of 29 digits) contains 22 as substring.
It is not a weakly prime, because it can be changed into another prime (211211102122187) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 105605551061073 + 105605551061074.
It is an arithmetic number, because the mean of its divisors is an integer number (105605551061074).
Almost surely, 2211211102122147 is an apocalyptic number.
211211102122147 is a deficient number, since it is larger than the sum of its proper divisors (1).
211211102122147 is an equidigital number, since it uses as much as digits as its factorization.
211211102122147 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 896, while the sum is 28.
Adding to 211211102122147 its reverse (741221201112112), we get a palindrome (952432303234259).
The spelling of 211211102122147 in words is "two hundred eleven trillion, two hundred eleven billion, one hundred two million, one hundred twenty-two thousand, one hundred forty-seven".
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