Base | Representation |
---|---|
bin | 110101000101110111… |
… | …0000000001011000111 |
3 | 101220021210000210111201 |
4 | 1222023232000023013 |
5 | 3331444342221210 |
6 | 124213231243331 |
7 | 11144232365416 |
oct | 1521356001307 |
9 | 356253023451 |
10 | 114013242055 |
11 | 44397559289 |
12 | 1a11a979547 |
13 | a99caa6a4b |
14 | 57381c347d |
15 | 2e745d653a |
hex | 1a8bb802c7 |
114013242055 has 4 divisors (see below), whose sum is σ = 136815890472. Its totient is φ = 91210593640.
The previous prime is 114013242023. The next prime is 114013242071. The reversal of 114013242055 is 550242310411.
It is a semiprime because it is the product of two primes, and also an emirpimes, since its reverse is a distinct semiprime: 550242310411 = 193 ⋅2850996427.
It is not a de Polignac number, because 114013242055 - 25 = 114013242023 is a prime.
It is a Duffinian number.
It is a congruent number.
It is an unprimeable number.
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 in 3 ways as a sum of consecutive naturals, for example, 11401324201 + ... + 11401324210.
It is an arithmetic number, because the mean of its divisors is an integer number (34203972618).
Almost surely, 2114013242055 is an apocalyptic number.
114013242055 is a deficient number, since it is larger than the sum of its proper divisors (22802648417).
114013242055 is an equidigital number, since it uses as much as digits as its factorization.
114013242055 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 22802648416.
The product of its (nonzero) digits is 4800, while the sum is 28.
Adding to 114013242055 its reverse (550242310411), we get a palindrome (664255552466).
The spelling of 114013242055 in words is "one hundred fourteen billion, thirteen million, two hundred forty-two thousand, fifty-five".
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