Base | Representation |
---|---|
bin | 10110100111011100… |
… | …01001010110110001 |
3 | 1011100012100201210210 |
4 | 23103232021112301 |
5 | 144331324232423 |
6 | 5324501414333 |
7 | 606614334132 |
oct | 132356112661 |
9 | 34305321723 |
10 | 12142024113 |
11 | 517094647a |
12 | 242a40a3a9 |
13 | 11b670026b |
14 | 83282bc89 |
15 | 4b0e76b93 |
hex | 2d3b895b1 |
12142024113 has 8 divisors (see below), whose sum is σ = 16193520576. Its totient is φ = 8092605200.
The previous prime is 12142024091. The next prime is 12142024151. The reversal of 12142024113 is 31142024121.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 12142024113 - 212 = 12142020017 is a prime.
It is a super-3 number, since 3×121420241133 (a number of 31 digits) contains 333 as substring.
It is a self number, because there is not a number n which added to its sum of digits gives 12142024113.
It is not an unprimeable number, because it can be changed into a prime (12142024183) by changing a digit.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 505698 + ... + 529163.
It is an arithmetic number, because the mean of its divisors is an integer number (2024190072).
Almost surely, 212142024113 is an apocalyptic number.
It is an amenable number.
12142024113 is a deficient number, since it is larger than the sum of its proper divisors (4051496463).
12142024113 is a wasteful number, since it uses less digits than its factorization.
12142024113 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1038775.
The product of its (nonzero) digits is 384, while the sum is 21.
Adding to 12142024113 its reverse (31142024121), we get a palindrome (43284048234).
The spelling of 12142024113 in words is "twelve billion, one hundred forty-two million, twenty-four thousand, one hundred thirteen".
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