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(2x+3y) is divisible by 17,then for what value of k; (9x+ky) is divisible by 17.Please give the solution.
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4(2x+3y)+(9x+5y) = 17(x+y). Since RHS is divisible by 17, and so is 4(2x+3y), we must have 9x+5y also divisible. Hence for any k such that k = 5+19m for any integer m, we have 9x+ky divisible by 17
gcd(2x+ 3y, 9x + ky ) = 17 or gcd(2x + 3y, (9x + ky) - 4(2x + 3y)) = 17 or gcd(2x + 3y, (x + (k-12) y) = 17 gcd ( 2x+ 3y - 2(x + k- 12y), (x + (k-12) y)) = 17 so 27y -2k y is divisible by 17 so 27 - 2k is divisible by 17 or k = 5
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