chess4math

Working backwards,

\(a_9 = 1 - 4 = -3 \)

\(a_8 = 4 - (-3) = 7\)

\(a_7 = -3-7=-10\)

\(a_6 = 7-(-10)=\boxed{17}\)

We can write the equation in slope intercept form (y = mx +b) and convert it to Ax+By=C form.

m is the slope which is equal to (1-5)/(9-(-2)) = -4/11

y = -4x/11 + b

Substituting a point, we have 1 = -36/11 + b

b = 1 + 36/11 = 47/11

y = -4x/11 + 47/11

11y = -4x + 47

4x + 11y = 47

Let x be the units digit. The value of the hexadecimal integer in base 10 is  \(2 \cdot 16^3 + 15 \cdot 16^2 +1 \cdot 16 +x = 12048+x\)

The smallest multiple of 19 greater than 12048 is 12065. In order for 12048 + x to be a multiple of 19, x has to be 17 which is not possible in hexadecimal. Therefore, it is not possible.