Base | Representation |
---|---|
bin | 1100001110010111110010… |
… | …1010010000101011011010 |
3 | 1202120221200110211100100012 |
4 | 3003211330222100223122 |
5 | 3230204224243403310 |
6 | 44330421434441522 |
7 | 2555040413651546 |
oct | 303457452205332 |
9 | 52527613740305 |
10 | 13441044122330 |
11 | 4312350aaa750 |
12 | 1610b6a5932a2 |
13 | 76663c794176 |
14 | 34679b30aa26 |
15 | 184974561c05 |
hex | c397ca90ada |
13441044122330 has 64 divisors (see below), whose sum is σ = 27549046204416. Its totient is φ = 4673755944000.
The previous prime is 13441044122329. The next prime is 13441044122339. The reversal of 13441044122330 is 3322144014431.
13441044122330 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is not an unprimeable number, because it can be changed into a prime (13441044122339) by changing a digit.
It is a polite number, since it can be written in 31 ways as a sum of consecutive naturals, for example, 7740635 + ... + 9316625.
It is an arithmetic number, because the mean of its divisors is an integer number (430453846944).
Almost surely, 213441044122330 is an apocalyptic number.
13441044122330 is a gapful number since it is divisible by the number (10) formed by its first and last digit.
13441044122330 is an abundant number, since it is smaller than the sum of its proper divisors (14108002082086).
It is a pseudoperfect number, because it is the sum of a subset of its proper divisors.
13441044122330 is a wasteful number, since it uses less digits than its factorization.
13441044122330 is an evil number, because the sum of its binary digits is even.
The sum of its prime factors is 1579403.
The product of its (nonzero) digits is 27648, while the sum is 32.
Adding to 13441044122330 its reverse (3322144014431), we get a palindrome (16763188136761).
The spelling of 13441044122330 in words is "thirteen trillion, four hundred forty-one billion, forty-four million, one hundred twenty-two thousand, three hundred thirty".
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