Base | Representation |
---|---|
bin | 10001101000110010000… |
… | …110110110001100110101 |
3 | 11021212102212011012001221 |
4 | 101220302012312030311 |
5 | 124324204441330031 |
6 | 2324443413350341 |
7 | 153365010125041 |
oct | 21506206661465 |
9 | 4255385135057 |
10 | 1212021433141 |
11 | 428019557802 |
12 | 176a936383b1 |
13 | 8a3a701505c |
14 | 4293ad25021 |
15 | 217da291a11 |
hex | 11a321b6335 |
1212021433141 has 2 divisors, whose sum is σ = 1212021433142. Its totient is φ = 1212021433140.
The previous prime is 1212021433099. The next prime is 1212021433159. The reversal of 1212021433141 is 1413341202121.
It is a strong prime.
It can be written as a sum of positive squares in only one way, i.e., 961486146916 + 250535286225 = 980554^2 + 500535^2 .
It is a cyclic number.
It is not a de Polignac number, because 1212021433141 - 219 = 1212020908853 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1212021433241) 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 as a sum of consecutive naturals, namely, 606010716570 + 606010716571.
It is an arithmetic number, because the mean of its divisors is an integer number (606010716571).
Almost surely, 21212021433141 is an apocalyptic number.
It is an amenable number.
1212021433141 is a deficient number, since it is larger than the sum of its proper divisors (1).
1212021433141 is an equidigital number, since it uses as much as digits as its factorization.
1212021433141 is an odious number, because the sum of its binary digits is odd.
The product of its (nonzero) digits is 1152, while the sum is 25.
Adding to 1212021433141 its reverse (1413341202121), we get a palindrome (2625362635262).
The spelling of 1212021433141 in words is "one trillion, two hundred twelve billion, twenty-one million, four hundred thirty-three thousand, one hundred forty-one".
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