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Question

If 4x17y=1 & x,y500. 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 A 29
4x17y=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+(n1)(4)
112=4n4
4n=116
n=1164 = 29
Hence, there can be 29 possible values.

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