Base | Representation |
---|---|
bin | 10100100010001101100… |
… | …100000011111111111111 |
3 | 11222220100200100111121121 |
4 | 110202031210003333333 |
5 | 141104440311022421 |
6 | 3000132255211411 |
7 | 203643655522501 |
oct | 24421544037777 |
9 | 4886320314547 |
10 | 1411124314111 |
11 | 4a44aa984116 |
12 | 1a959a816b67 |
13 | a30b7508c65 |
14 | 4c427bd9d71 |
15 | 26a8eaa7d41 |
hex | 1488d903fff |
1411124314111 has 2 divisors, whose sum is σ = 1411124314112. Its totient is φ = 1411124314110.
The previous prime is 1411124314027. The next prime is 1411124314133. The reversal of 1411124314111 is 1114134211141.
It is a strong prime.
It is a cyclic number.
It is not a de Polignac number, because 1411124314111 - 27 = 1411124313983 is a prime.
It is a congruent number.
It is not a weakly prime, because it can be changed into another prime (1411124314811) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (23) of ones.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 705562157055 + 705562157056.
It is an arithmetic number, because the mean of its divisors is an integer number (705562157056).
Almost surely, 21411124314111 is an apocalyptic number.
1411124314111 is a deficient number, since it is larger than the sum of its proper divisors (1).
1411124314111 is an equidigital number, since it uses as much as digits as its factorization.
1411124314111 is an odious number, because the sum of its binary digits is odd.
The product of its digits is 384, while the sum is 25.
Adding to 1411124314111 its reverse (1114134211141), we get a palindrome (2525258525252).
The spelling of 1411124314111 in words is "one trillion, four hundred eleven billion, one hundred twenty-four million, three hundred fourteen thousand, one hundred eleven".
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