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Grade: 11
        (2x+3y) is divisible by 17,then for what value of k; (9x+ky) is divisible by 17.
Please give the solution.
6 years ago

Answers : (3)

Shrey
49 Points
							Please reply fast
						
6 years ago
mycroft holmes
272 Points
							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
						
6 years ago
Sher Mohammad
IIT Delhi
askIITians Faculty
174 Points
							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
6 years ago
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