引言
gcd 有目前几条性质:
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\(a \cdot b = lcm(a,b) \cdot gcd(a,b)\)
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\(gcd(a,b) = gcd(b,a-b)\)
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\(gcd(a,b) = gcd(b,a+b)\)
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\(gcd(a,b) = gcd(b,a \% b)\)
性质1
\(a \cdot b = lcm(a,b) \cdot gcd(a,b)\)
证明:
设 \(d = gcd(a,b)\),则有 \(a= a_0 \cdot d\),\(b = b_0 \cdot d\)
由于 \(gcd(a_0,b_0) = 1\)
并且 \(lcm(a_0,b_0) = a_0 \cdot b_0\)
则有 \(a \cdot b = a_0 \cdot b_0 \cdot d = lcm(a,b) \cdot gcd(a,b)\)
性质2
\(gcd(a,b) = gcd(a,a-b)\)
设 \(d = gcd(a,b)\),则有 \(a = md\),\(b = nd\),且 \(gcd(m,n)= 1\)
则有 \(gcd(a,a-b) = gcd(md,md-nd) = d * gcd(m,m-n) = d\)
所以 \(gcd(a,b) = gcd(a,a-b)\)
性质3
\(gcd(a,b) = gcd(a,a+b)\)
设 \(d = gcd(a,b)\),则有 \(a = md\),\(b = nd\),且 \(gcd(m,n)= 1\)
则有 \(gcd(a,a+b) = gcd(md,md+nd) = d * (m,m+n) = d\)
所以 \(gcd(a,b) = gcd(a,a+b)\)
性质4
\(gcd(a,b) = gcd(b,a\%b)\)
设 \(a = k \cdot b + r\),则有 \(r = a - k \cdot b\)
设 \(gcd(a,b) = d\)
则由 \(r = a - k \cdot b = n \cdot d - k \cdot m \cdot d = (n - k \cdot m) \cdot d\)
所以 d 是 r 的因数,因此 d 是 b 和 r 的公因数
由于 \(r = a \% b\),则可证 \(gcd(a,b) = gcd(b,a%b)\)