Base | Representation |
---|---|
bin | 10110011000000111… |
… | …00000000010110101 |
3 | 1011000020100011122112 |
4 | 23030003200002311 |
5 | 144100412414331 |
6 | 5304024540405 |
7 | 603463152311 |
oct | 131403400265 |
9 | 34006304575 |
10 | 12013404341 |
11 | 5105287544 |
12 | 23b3321705 |
13 | 1195b78aa3 |
14 | 81d70ab41 |
15 | 4a4a1c32b |
hex | 2cc0e00b5 |
12013404341 has 8 divisors (see below), whose sum is σ = 12658392480. Its totient is φ = 11369686608.
The previous prime is 12013404323. The next prime is 12013404401. The reversal of 12013404341 is 14340431021.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is a de Polignac number, because none of the positive numbers 2k-12013404341 is a prime.
It is a Duffinian number.
It is a Curzon number.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (12013400341) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 298151 + ... + 336036.
It is an arithmetic number, because the mean of its divisors is an integer number (1582299060).
Almost surely, 212013404341 is an apocalyptic number.
It is an amenable number.
12013404341 is a deficient number, since it is larger than the sum of its proper divisors (644988139).
12013404341 is an equidigital number, since it uses as much as digits as its factorization.
12013404341 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 635203.
The product of its (nonzero) digits is 1152, while the sum is 23.
Adding to 12013404341 its reverse (14340431021), we get a palindrome (26353835362).
The spelling of 12013404341 in words is "twelve billion, thirteen million, four hundred four thousand, three hundred forty-one".
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