If 4x−17y=1 & x,y≤500. find how many positive integer solutions are possible ?
A
29
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B
28
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C
27
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D
Can not be determined
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Solution
The correct option is A29 4x−17y=1 x=1+17y4 x is an integer when y=3,7,11,15,...115 Thus, number of values of y form an AP.
Maximum value of y can be 115 else x will be greater than 500 Hence, using the nth term formula of an AP, 115=3+(n−1)(4) 112=4n−4 4n=116 n=1164 = 29 Hence, there can be 29 possible values.