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26 Sep 2022, 09:25
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solve once again the classic linear Diophantine Equation: 2x+9y=57, when x,y∈Z .
Step 1: Determine the GCD of 2 and 9 which is gcd(2,9)=1.
Step 2: Since 1∣57 , we will continue to find the solution
Step 3: Find a particular solution to 2x+9y=57, x,y∈Z .
Since 2(6)+9(5)=57 , x=6 and y=5 is a particular solution.
Step 4: Let u=x−6 and v=y-5.
Then 2u+9v=2(x−6)+9(y-5)=0.
Step 5: Solve 2u+9v=0
The general solutions is u=−9m and v=2m, m∈Z .
Step 6: x−6=−9m and y-5=2m, m∈Z .
Hence, the general solutions is x=−9m+6, y=2m+5, m∈Z . We want now to find the min value of (x+y)^2
Substituting we have (-9m+6+2m+5)^2 = (11-7m)^2 .
The first derivative is 2(11-7m)(-7)= -14(11-7m). Minimum exist where -14(11-7m)=0 => 7m=11=> m=1.571. But m∈Z , so m=2 . For m=2 the derivative is positive , so we have minimum.
For m=2 , x=-12 and y=9 . The value of (x+y)^2=(-12+9)^2=9
Answer is B.
Step 1: Determine the GCD of 2 and 9 which is gcd(2,9)=1.
Step 2: Since 1∣57 , we will continue to find the solution
Step 3: Find a particular solution to 2x+9y=57, x,y∈Z .
Since 2(6)+9(5)=57 , x=6 and y=5 is a particular solution.
Step 4: Let u=x−6 and v=y-5.
Then 2u+9v=2(x−6)+9(y-5)=0.
Step 5: Solve 2u+9v=0
The general solutions is u=−9m and v=2m, m∈Z .
Step 6: x−6=−9m and y-5=2m, m∈Z .
Hence, the general solutions is x=−9m+6, y=2m+5, m∈Z . We want now to find the min value of (x+y)^2
Substituting we have (-9m+6+2m+5)^2 = (11-7m)^2 .
The first derivative is 2(11-7m)(-7)= -14(11-7m). Minimum exist where -14(11-7m)=0 => 7m=11=> m=1.571. But m∈Z , so m=2 . For m=2 the derivative is positive , so we have minimum.
For m=2 , x=-12 and y=9 . The value of (x+y)^2=(-12+9)^2=9
Answer is B.
24 Sep 2025, 07:37
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