Base | Representation |
---|---|
bin | 1010010010011110000110… |
… | …1100000101010110010101 |
3 | 1111001110101001211102110211 |
4 | 2210213201230011112111 |
5 | 2440320331420221323 |
6 | 40020505552445421 |
7 | 2245203456451606 |
oct | 244474154052625 |
9 | 44043331742424 |
10 | 11312435320213 |
11 | 3671638907355 |
12 | 1328513849871 |
13 | 6409b16444b1 |
14 | 2b174c051dad |
15 | 1493e092770d |
hex | a49e1b05595 |
11312435320213 has 2 divisors, whose sum is σ = 11312435320214. Its totient is φ = 11312435320212.
The previous prime is 11312435320189. The next prime is 11312435320339. The reversal of 11312435320213 is 31202353421311.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 8996508338724 + 2315926981489 = 2999418^2 + 1521817^2 .
It is a cyclic number.
It is not a de Polignac number, because 11312435320213 - 217 = 11312435189141 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (11312435320013) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5656217660106 + 5656217660107.
It is an arithmetic number, because the mean of its divisors is an integer number (5656217660107).
Almost surely, 211312435320213 is an apocalyptic number.
It is an amenable number.
11312435320213 is a deficient number, since it is larger than the sum of its proper divisors (1).
11312435320213 is an equidigital number, since it uses as much as digits as its factorization.
11312435320213 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 12960, while the sum is 31.
Adding to 11312435320213 its reverse (31202353421311), we get a palindrome (42514788741524).
The spelling of 11312435320213 in words is "eleven trillion, three hundred twelve billion, four hundred thirty-five million, three hundred twenty thousand, two hundred thirteen".
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