Base | Representation |
---|---|
bin | 1100110101010100010010… |
… | …1101011111101010010011 |
3 | 1211221220200112210001100121 |
4 | 3031111010231133222103 |
5 | 3322140011302003321 |
6 | 50002033320131111 |
7 | 2654264636004553 |
oct | 315250455375223 |
9 | 54856615701317 |
10 | 14110120344211 |
11 | 45500830a2121 |
12 | 16ba776906a97 |
13 | 7b476a368638 |
14 | 36ad0d61b163 |
15 | 197083868441 |
hex | cd544b5fa93 |
14110120344211 has 2 divisors, whose sum is σ = 14110120344212. Its totient is φ = 14110120344210.
The previous prime is 14110120344209. The next prime is 14110120344217. The reversal of 14110120344211 is 11244302101141.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 14110120344211 - 21 = 14110120344209 is a prime.
It is a super-2 number, since 2×141101203442112 (a number of 27 digits) contains 22 as substring.
Together with 14110120344209, it forms a pair of twin primes.
It is not a weakly prime, because it can be changed into another prime (14110120344217) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7055060172105 + 7055060172106.
It is an arithmetic number, because the mean of its divisors is an integer number (7055060172106).
Almost surely, 214110120344211 is an apocalyptic number.
14110120344211 is a deficient number, since it is larger than the sum of its proper divisors (1).
14110120344211 is an equidigital number, since it uses as much as digits as its factorization.
14110120344211 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 768, while the sum is 25.
Adding to 14110120344211 its reverse (11244302101141), we get a palindrome (25354422445352).
The spelling of 14110120344211 in words is "fourteen trillion, one hundred ten billion, one hundred twenty million, three hundred forty-four thousand, two hundred eleven".
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