Question

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1 (2 points). Prove that the following equation has an integer solution for \( x \) and \( y \). \[ 41 x-132 y=41132 \]

Answer 1

There are multiple integer solutions possible. I found two solutions.

X = 1132 & Y = 40. X = 76 & Y = -288.

Explanation:

41 x - 132 y = 41132

41 (x - 3 y - 1003) - 9 y = 9

41 (x - 3 y - 1003) = 9 (y + 1)

As 41 & 9 are relatively prime,

y + 1 = 41 => y = 40.

x - 3 y - 1003 = 9 => x = 1132.

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41 x = 132 (y + 311) + 80

41 [x - 3 (y+311)] - 9 (y+311) = 80

5 [x - 3 (y+311)] - 9 [ 4{x - 3(y+311)} - (y+ 311) ] = 80

5 [x - 3 (y+311)] - 9 [ 4x - 13(y+311) ] = 80

One possible solution :

4x - 13(y+311) = 5, => 4 x - 13 y = 4048

and x - 3(y + 311) = 7, => x - 3 y = 940

Solving these equations we get x = 76 & y = -288.