Question

The number of positive integral values of a such that the lines x – 4y = 1 and ax + 3y = 1 intersect at an integral point, (integral point is a point both of whose coordinates are integers), is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

1

Answer 2

For a = 1, we have x = y = 1 for a > 1

a(4y + 1 ) = 1, 1 – 3y so y = $\frac { - ( a - 1 ) } { 4 a + 3 }$ < 0

But a – 1 < 4a + 3, since a is positive, so y > – 1 (contradiction)

a = 1 only