Official solution
6.35Cr
= (k2 - 3). 36Cr + 1
\[ \Rightarrow \] 6.35Cr
= (k2 - 3).\[{{36} \over {r + 1}}\]35Cr
\[ \Rightarrow \] k2 - 3 = \[{{r + 1} \over 6}\]
Possible values of r for integral values of k, are
r = 5, 35
number of ordered pairs are 4
(5, 2), (5, --2), (35, 3), (35, 3)