Base | Representation |
---|---|
bin | 10010111111100110100… |
… | …100001101110010001011 |
3 | 11121210001122121221000111 |
4 | 102333212210031302023 |
5 | 132341114244002021 |
6 | 2435342010415151 |
7 | 163205101236535 |
oct | 22774644156213 |
9 | 4553048557014 |
10 | 1305243344011 |
11 | 463606927152 |
12 | 190b6b4314b7 |
13 | 961124a7c69 |
14 | 47261ab2855 |
15 | 23e443e88e1 |
hex | 12fe690dc8b |
1305243344011 has 2 divisors, whose sum is σ = 1305243344012. Its totient is φ = 1305243344010.
The previous prime is 1305243343973. The next prime is 1305243344081. The reversal of 1305243344011 is 1104433425031.
It is a weak prime.
It is an emirp because it is prime and its reverse (1104433425031) is a distict prime.
It is a cyclic number.
It is not a de Polignac number, because 1305243344011 - 29 = 1305243343499 is a prime.
It is not a weakly prime, because it can be changed into another prime (1305243344081) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 652621672005 + 652621672006.
It is an arithmetic number, because the mean of its divisors is an integer number (652621672006).
Almost surely, 21305243344011 is an apocalyptic number.
1305243344011 is a deficient number, since it is larger than the sum of its proper divisors (1).
1305243344011 is an equidigital number, since it uses as much as digits as its factorization.
1305243344011 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 17280, while the sum is 31.
Adding to 1305243344011 its reverse (1104433425031), we get a palindrome (2409676769042).
The spelling of 1305243344011 in words is "one trillion, three hundred five billion, two hundred forty-three million, three hundred forty-four thousand, eleven".
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