// Numbas version: finer_feedback_settings {"name": "Modular Arithmetic: Greatest Common Divisor", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Modular Arithmetic: Greatest Common Divisor", "tags": [], "metadata": {"description": "

Introduces the GCD and a simple form of the Euclidian Algorithm

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A number $n$ divides $a$ when $a = nq$ for some $q$.

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Any number which divides both $a$ and $b$ is called a common divisor of $a$ and $b$, the largest common divisor is called the greatest common divisor and is denoted $\\gcd(a,b)$.

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Every number divides zero, so the largest number to divide $\\var{a1}$ is the greatest common divisor of $0$ and $\\var{a1}$. This means that $\\gcd(\\var{a1},0) = \\var{a1}$.

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The Euclidian Algorithm presented here is a variation on the usual presentation. Another common method keeps track of the quotients:

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$ \\var{gcd*(1+b*(1+a))} = \\var{b}\\times (\\var{gcd*(1+a)}) + \\var{a1} $

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$ \\var{gcd*(1+a)} = \\var{gcd*(1+a)/a1}\\times (\\var{a1}) + 0 $

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to write the $\\gcd$ in the form of Bézout's identity

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$ \\var{gcd*(1+b*(1+a))} - \\var{b}\\times (\\var{gcd*(1+a)}) = \\var{a1}$.

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As a consequence of these definitions, every number $n$ divides zero because $0 = n\\times 0$.

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So $\\gcd(\\var{big},0)$ is

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What is $\\gcd(\\var{gcd*(1+b*(1+a))},\\var{gcd*(1+a)})$?

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The procedure is to use the identity

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 $\\gcd(a,b) = \\gcd(a\\mod b, b)$.

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to simplify the question.

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Compute $\\var{gcd*(1+b*(1+a))} \\mod{\\var{gcd*(1+a)}}$.

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Hence $\\gcd(\\var{gcd*(1+b*(1+a))},\\var{gcd*(1+a)}) = \\gcd(\\var{a1},\\var{gcd*(1+a)})$. We can simplify this again using the identity

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 $\\gcd(a,b) = \\gcd(a, b\\mod a)$.

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Compute $\\var{gcd*(a+1)} \\mod{\\var{a1}} $.

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By re-applying these identities, the question has been simplified to something managable:

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$\\gcd(\\var{gcd*(1+b*(1+a))},\\var{gcd*(1+a)}) = \\gcd(\\var{a1},\\var{gcd*(1+a)}) = \\gcd(\\var{a1}, 0)$.

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