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Albert1
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find: $gcd(2002+2,2002^2+2,2002^3+2,---------------,2002^{2002}+2)$
Albert said:find: $gcd(2002+2,2002^2+2,2002^3+2,---------------,2002^{2002}+2)$
The purpose of calculating gcd(2002+2,2002^2+2,...,2002^2002+2) is to find the greatest common divisor of all the given numbers. This can help in simplifying fractions, finding common factors, and solving certain mathematical problems.
To calculate gcd(2002+2,2002^2+2,...,2002^2002+2), you can use the Euclidean algorithm which involves repeatedly dividing the larger number by the smaller number until the remainder is 0. The last non-zero remainder is the gcd.
Yes, gcd(2002+2,2002^2+2,...,2002^2002+2) can be greater than 1. In fact, if all the given numbers are even, then the gcd will be at least 2.
Yes, there are faster algorithms such as the binary GCD algorithm and the Lehmer-Euclid algorithm that can be used to calculate gcd(2002+2,2002^2+2,...,2002^2002+2) for a large set of numbers.
The significance of using 2002 as the base number is that it is a composite number, which means it has more than two factors. This allows for a greater range of possible gcd values, making the calculation more interesting and challenging.