Base | Representation |
---|---|
bin | 1100111011111001100110… |
… | …0011110111101111010011 |
3 | 1212100201121011121222011122 |
4 | 3032332121203313233103 |
5 | 3331013120014100301 |
6 | 50130015345032455 |
7 | 2665410323006621 |
oct | 316763143675723 |
9 | 55321534558148 |
10 | 14223213034451 |
11 | 4594037a61183 |
12 | 171867932572b |
13 | 7c2321501365 |
14 | 37259b454511 |
15 | 199ea221381b |
hex | cef998f7bd3 |
14223213034451 has 2 divisors, whose sum is σ = 14223213034452. Its totient is φ = 14223213034450.
The previous prime is 14223213034411. The next prime is 14223213034453. The reversal of 14223213034451 is 15443031232241.
It is a strong prime.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-14223213034451 is a prime.
Together with 14223213034453, 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 (14223213034453) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (29) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 7111606517225 + 7111606517226.
It is an arithmetic number, because the mean of its divisors is an integer number (7111606517226).
Almost surely, 214223213034451 is an apocalyptic number.
14223213034451 is a deficient number, since it is larger than the sum of its proper divisors (1).
14223213034451 is an equidigital number, since it uses as much as digits as its factorization.
14223213034451 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 69120, while the sum is 35.
Adding to 14223213034451 its reverse (15443031232241), we get a palindrome (29666244266692).
The spelling of 14223213034451 in words is "fourteen trillion, two hundred twenty-three billion, two hundred thirteen million, thirty-four thousand, four hundred fifty-one".
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