Base | Representation |
---|---|
bin | 11001000001001100011011… |
… | …001110110110011101000001 |
3 | 112102121001222120102120212001 |
4 | 121001030123032312131001 |
5 | 103410241020323042423 |
6 | 1030004322034104001 |
7 | 32114431603246012 |
oct | 3101143316663501 |
9 | 472531876376761 |
10 | 110033224034113 |
11 | 32072912452826 |
12 | 1041121411b001 |
13 | 495211c59b743 |
14 | 1d258ca62d009 |
15 | cac33b3d5cad |
hex | 64131b3b6741 |
110033224034113 has 2 divisors, whose sum is σ = 110033224034114. Its totient is φ = 110033224034112.
The previous prime is 110033224034101. The next prime is 110033224034147. The reversal of 110033224034113 is 311430422330011.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 109286785057024 + 746438977089 = 10454032^2 + 863967^2 .
It is a cyclic number.
It is not a de Polignac number, because 110033224034113 - 233 = 110024634099521 is a prime.
It is not a weakly prime, because it can be changed into another prime (110033224032113) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 55016612017056 + 55016612017057.
It is an arithmetic number, because the mean of its divisors is an integer number (55016612017057).
Almost surely, 2110033224034113 is an apocalyptic number.
It is an amenable number.
110033224034113 is a deficient number, since it is larger than the sum of its proper divisors (1).
110033224034113 is an equidigital number, since it uses as much as digits as its factorization.
110033224034113 is an evil number, because the sum of its binary digits is even.
The product of its (nonzero) digits is 5184, while the sum is 28.
Adding to 110033224034113 its reverse (311430422330011), we get a palindrome (421463646364124).
The spelling of 110033224034113 in words is "one hundred ten trillion, thirty-three billion, two hundred twenty-four million, thirty-four thousand, one hundred thirteen".
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