Base | Representation |
---|---|
bin | 1100000010010100101010… |
… | …0100100100110110111001 |
3 | 1201212011101101002201220122 |
4 | 3000211022210210312321 |
5 | 3213311311101403423 |
6 | 44051345252303025 |
7 | 2534061660232562 |
oct | 300451244446671 |
9 | 51764341081818 |
10 | 13234045341113 |
11 | 4242588647643 |
12 | 1598a1b011a75 |
13 | 74cc714a7a3c |
14 | 33a761909169 |
15 | 17e3ab9648c8 |
hex | c094a924db9 |
13234045341113 has 16 divisors (see below), whose sum is σ = 13827469641216. Its totient is φ = 12651071598720.
The previous prime is 13234045341089. The next prime is 13234045341119. The reversal of 13234045341113 is 31114354043231.
It is a cyclic number.
It is not a de Polignac number, because 13234045341113 - 214 = 13234045324729 is a prime.
It is a Duffinian number.
It is not an unprimeable number, because it can be changed into a prime (13234045341119) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (19) of ones.
It is a polite number, since it can be written in 15 ways as a sum of consecutive naturals, for example, 18265445 + ... + 18976157.
It is an arithmetic number, because the mean of its divisors is an integer number (864216852576).
Almost surely, 213234045341113 is an apocalyptic number.
It is an amenable number.
13234045341113 is a deficient number, since it is larger than the sum of its proper divisors (593424300103).
13234045341113 is an equidigital number, since it uses as much as digits as its factorization.
13234045341113 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 718064.
The product of its (nonzero) digits is 51840, while the sum is 35.
Adding to 13234045341113 its reverse (31114354043231), we get a palindrome (44348399384344).
The spelling of 13234045341113 in words is "thirteen trillion, two hundred thirty-four billion, forty-five million, three hundred forty-one thousand, one hundred thirteen".
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