Base | Representation |
---|---|
bin | 1100001110011100100111… |
… | …1001101010010110000011 |
3 | 1202121002000210021000111221 |
4 | 3003213021321222112003 |
5 | 3230214403020132001 |
6 | 44331154210015511 |
7 | 2555114500534063 |
oct | 303471171522603 |
9 | 52532023230457 |
10 | 13442340005251 |
11 | 4312956545773 |
12 | 1611270577597 |
13 | 7667b80ba1b1 |
14 | 3468814793a3 |
15 | 1849ed1dc1a1 |
hex | c39c9e6a583 |
13442340005251 has 2 divisors, whose sum is σ = 13442340005252. Its totient is φ = 13442340005250.
The previous prime is 13442340005243. The next prime is 13442340005287. The reversal of 13442340005251 is 15250004324431.
13442340005251 is digitally balanced in base 2, because in such base it contains all the possibile digits an equal number of times.
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 13442340005251 - 23 = 13442340005243 is a prime.
It is not a weakly prime, because it can be changed into another prime (13442340005231) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 6721170002625 + 6721170002626.
It is an arithmetic number, because the mean of its divisors is an integer number (6721170002626).
Almost surely, 213442340005251 is an apocalyptic number.
13442340005251 is a deficient number, since it is larger than the sum of its proper divisors (1).
13442340005251 is an equidigital number, since it uses as much as digits as its factorization.
13442340005251 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 57600, while the sum is 34.
Adding to 13442340005251 its reverse (15250004324431), we get a palindrome (28692344329682).
The spelling of 13442340005251 in words is "thirteen trillion, four hundred forty-two billion, three hundred forty million, five thousand, two hundred fifty-one".
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