$Total value of initial configeration = 13 hamper \times \$12 = \$156$

$Again total value of new configeration = 12 hamper \times \$13=\$156$

$\therefore$The value of gift certificates will not be changed.

$$\begin{gathered} Now Susan started with 13 hampers. \\ Susan has 13 piece of \$3 worth gift \end{gathered}$$

$$\begin{gathered} \therefore Susan has 13 piece of \$4 worth gift \\ \therefore Susan has 13 piece of \$5 worth gift \\ \therefore Possible set \{5, 5, 3\} \{4, 4, 5\} , (3, 3, 3, 4) in which \$5 contains in first two of them and the total \end{gathered}$$

number is 13.

Now, let Susan creates

x hampers of {3, 3, 3, 4}

y hampers of {4,4,5}

And z hampers of {5, 5, 3}

Number of $3 gift will be, 3x + z =13.....(i)

$\therefore Number of \$4 gift will be, x + 2y =13.....\left(ii\right)$

$$\begin{gathered} \therefore Number of \$5 gift will be, y + 2z = 13 \\ Now, multiplying equation \left(iii\right) by 2 and subtracting \left(ii\right) from \left(iii\right) we get , \end{gathered}$$

$$\begin{gathered} \frac{2y+4z=26 \\ x+2y=13 \\ - - - }{4z-x=13...............\left(iv\right)} \end{gathered}$$

$$\begin{gathered} Again, multiplying equation \left(i\right) by 4 and subtracting equation \left(iv\right) from \left(i\right) we get \\ \frac{12x+4z=52 \\ 4z-x=13}{13x=39} \\ \therefore x=3 \\ Here, x is the number of gift hampers without \$5 certificates. \\ \therefore 3 gift hampers did not contain \$5 gift certificates. \end{gathered}$$