Base | Representation |
---|---|
bin | 1010010010010101110100… |
… | …0011100111111001011011 |
3 | 1111001020121222201121102112 |
4 | 2210211131003213321123 |
5 | 2440301302242340212 |
6 | 40015501103022535 |
7 | 2245064356541336 |
oct | 244453503477133 |
9 | 44036558647375 |
10 | 11310210121307 |
11 | 36706a6836499 |
12 | 1327bb259aa4b |
13 | 64071862c256 |
14 | 2b15ba6d561d |
15 | 1493103ce422 |
hex | a495d0e7e5b |
11310210121307 has 2 divisors, whose sum is σ = 11310210121308. Its totient is φ = 11310210121306.
The previous prime is 11310210121283. The next prime is 11310210121319. The reversal of 11310210121307 is 70312101201311.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 11310210121307 - 232 = 11305915154011 is a prime.
It is not a weakly prime, because it can be changed into another prime (11310210121367) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5655105060653 + 5655105060654.
It is an arithmetic number, because the mean of its divisors is an integer number (5655105060654).
Almost surely, 211310210121307 is an apocalyptic number.
11310210121307 is a deficient number, since it is larger than the sum of its proper divisors (1).
11310210121307 is an equidigital number, since it uses as much as digits as its factorization.
11310210121307 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 252, while the sum is 23.
Adding to 11310210121307 its reverse (70312101201311), we get a palindrome (81622311322618).
The spelling of 11310210121307 in words is "eleven trillion, three hundred ten billion, two hundred ten million, one hundred twenty-one thousand, three hundred seven".
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