Base | Representation |
---|---|
bin | 1100111011101110000011… |
… | …0111010000011001100011 |
3 | 1212100102121012212112110212 |
4 | 3032323200313100121203 |
5 | 3330440243232044420 |
6 | 50124344053044335 |
7 | 2665240455322511 |
oct | 316734067203143 |
9 | 55312535775425 |
10 | 14220114331235 |
11 | 459279790562a |
12 | 1717b536266ab |
13 | 7c1c4953b667 |
14 | 372385ab74b1 |
15 | 199d701743c5 |
hex | ceee0dd0663 |
14220114331235 has 16 divisors (see below), whose sum is σ = 17962482823200. Its totient is φ = 10777209924096.
The previous prime is 14220114331211. The next prime is 14220114331237. The reversal of 14220114331235 is 53213341102241.
It is a happy number.
It is a cyclic number.
It is not a de Polignac number, because 14220114331235 - 214 = 14220114314851 is a prime.
It is not an unprimeable number, because it can be changed into a prime (14220114331237) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 6703817 + ... + 8566293.
It is an arithmetic number, because the mean of its divisors is an integer number (1122655176450).
Almost surely, 214220114331235 is an apocalyptic number.
14220114331235 is a deficient number, since it is larger than the sum of its proper divisors (3742368491965).
14220114331235 is a wasteful number, since it uses less digits than its factorization.
14220114331235 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 1942870.
The product of its (nonzero) digits is 17280, while the sum is 32.
Adding to 14220114331235 its reverse (53213341102241), we get a palindrome (67433455433476).
The spelling of 14220114331235 in words is "fourteen trillion, two hundred twenty billion, one hundred fourteen million, three hundred thirty-one thousand, two hundred thirty-five".
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