Base | Representation |
---|---|
bin | 111111001011001010111… |
… | …101100011100101001001 |
3 | 120101000201012110001200201 |
4 | 333023022331203211021 |
5 | 1032112011424244423 |
6 | 13122213103132201 |
7 | 625436003005342 |
oct | 77131275434511 |
9 | 16330635401621 |
10 | 4341322103113 |
11 | 1424165170606 |
12 | 5a1463ba4061 |
13 | 256500883318 |
14 | 11019a4850c9 |
15 | 77ddb25a2ad |
hex | 3f2caf63949 |
4341322103113 has 2 divisors, whose sum is σ = 4341322103114. Its totient is φ = 4341322103112.
The previous prime is 4341322103101. The next prime is 4341322103179. The reversal of 4341322103113 is 3113012231434.
It is a weak prime.
It can be written as a sum of positive squares in only one way, i.e., 4099600414009 + 241721689104 = 2024747^2 + 491652^2 .
It is a cyclic number.
It is not a de Polignac number, because 4341322103113 - 217 = 4341321972041 is a prime.
It is not a weakly prime, because it can be changed into another prime (4341322106113) by changing a digit.
It is a polite number, since it can be written as a sum of consecutive naturals, namely, 2170661051556 + 2170661051557.
It is an arithmetic number, because the mean of its divisors is an integer number (2170661051557).
Almost surely, 24341322103113 is an apocalyptic number.
It is an amenable number.
4341322103113 is a deficient number, since it is larger than the sum of its proper divisors (1).
4341322103113 is an equidigital number, since it uses as much as digits as its factorization.
4341322103113 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 4341322103113 its reverse (3113012231434), we get a palindrome (7454334334547).
The spelling of 4341322103113 in words is "four trillion, three hundred forty-one billion, three hundred twenty-two million, one hundred three thousand, one hundred thirteen".
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