, 2022

Factoring, $2022 = 2 $\cdot$ 1011 = 2
$\cdot 3 \cdot$ 337$. (It turns out that 337 is a prime number,
though this fact is not needed here.)

Therefore, $2022 = 2 \cdot 1011$ and
$2022 = 3 \cdot 674$ and $2022 = 6 \cdot 337$.

Thus, the three ordered pairs are $(a,b) =
(2, 1011), (3, 674), (6, 337)$.
Manipulating algebraically, the following equations are
equivalent: $$\begin{aligned} \dfrac{2c+1}{2d+1} & = \dfrac{1}{17}\\ 17(2c+1) & = 2d + 1 \\ 34c + 17 & = 2d + 1 \\ 34c + 16 & = 2d \\ d & = 17c + 8\end{aligned}$$ Since $c$ is an integer with $c > 0$, then $c \geq 1$, which means that $17c + 8 \geq 25$.

Therefore, the smallest possible value of $d$ is $d=25$.

Note that, when $d = 25$, we obtain
$c=1$ and so $\$\dfrac{2c+1}{2d+1} = \dfrac{3}{51} = \dfrac{1}{17}\$.$
Solution 1

When $x = -5$, the left side of the
equation equals 0.

This means that when $x = -5$, the
right side of the equation must equal 0 as well.

Thus, $(-5)^2 + 3(-5) + t = 0$ and so
$25 - 15 + t = 0$ or $t = -10$.

Solution 2

Expanding the left side, we obtain $$(px+r)(x+5) = px^2 + rx + 5px + 5r$$ Since
this is equal to $x^2 + 3x + t$ for
all real numbers, then the coefficients of the two quadratic expressions
must be the same.

Comparing coefficients of $x^2$, we
obtain $p = 1$.

This means that $$x^2 + rx + 5x + 5r = x^2 + 3x + t$$ Comparing coefficients of $x$, we obtain $r+5 = 3$ and so $r = -2$.

This means that $$x^2 + 3x - 10 = x^2 + 3x + t$$ Comparing constant terms, we obtain $t = -10$.