Base | Representation |
---|---|
bin | 1010001100101100000110… |
… | …0111111011110111111111 |
3 | 1110200222000010120120122211 |
4 | 2203023001213323313333 |
5 | 2432203423424122421 |
6 | 35503122201121251 |
7 | 2235056252566606 |
oct | 243130147736777 |
9 | 43628003516584 |
10 | 11213113114111 |
11 | 36335000338a7 |
12 | 1311214a89227 |
13 | 63451339a83c |
14 | 2aaa09028a3d |
15 | 146a2ae759e1 |
hex | a32c19fbdff |
11213113114111 has 2 divisors, whose sum is σ = 11213113114112. Its totient is φ = 11213113114110.
The previous prime is 11213113114097. The next prime is 11213113114183. The reversal of 11213113114111 is 11141131131211.
It is a m-pointer prime, because the next prime (11213113114183) can be obtained adding 11213113114111 to its product of digits (72).
It is a weak prime.
It is a cyclic number.
It is not a de Polignac number, because 11213113114111 - 223 = 11213104725503 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (11213113115111) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 5606556557055 + 5606556557056.
It is an arithmetic number, because the mean of its divisors is an integer number (5606556557056).
Almost surely, 211213113114111 is an apocalyptic number.
11213113114111 is a deficient number, since it is larger than the sum of its proper divisors (1).
11213113114111 is an equidigital number, since it uses as much as digits as its factorization.
11213113114111 is an evil number, because the sum of its binary digits is even.
The product of its digits is 72, while the sum is 22.
Adding to 11213113114111 its reverse (11141131131211), we get a palindrome (22354244245322).
The spelling of 11213113114111 in words is "eleven trillion, two hundred thirteen billion, one hundred thirteen million, one hundred fourteen thousand, one hundred eleven".
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