0

Question: The point of intersection of $4x+5y=26$ and $y=kx+2$ has integral coordinates. What is the number of integral values that $k$ can take?

As per me the answer should be that $k$ can take $3$ integral values which will make coordinates of the point of intersection of the $2$ equations integer.

How I did this is by equating the equations. Then I got $x = \frac{16}{4+5k}$

Now $16$ has $5$ positive factors $(1, 2, 4, 8, 16)$ and $x$ can even take negative values

Then $5k+4$ can have values ranging from $1, 2, 4, 8, 16$ and $-1, -2, -4, -8, -16$

We will see that $k$ has an integral at $5k+4 = 4, -1, -16$ which makes $3$ values for $k$ Is this correct since the answer is given as $1$ which is not true has I got $k = 0$ and $-1$ which satisfies the conditions and the true answer should be $3$?

Plus if this is correct, any shorter way for this?

nmasanta
  • 9,222
Ayush Sambher
  • 187
  • This looks correct to me. I expect they forgot the negative factors. It's not too long: Find x; you need the factors of 16; pick the ones that work. – Empy2 Jun 14 '21 at 16:03
  • This is correct. The only slight shortening I see is to note that if a factor $f$ is of the form $5k+4$, then $f\equiv4\pmod5$, which makes it a bit easier to pick out the factors that work. It's not much of an improvement, though, and I think you've done fine. – saulspatz Jun 14 '21 at 16:07
  • Thank You guys! – Ayush Sambher Jun 14 '21 at 16:13

0 Answers0