Let's define the variables for the number of tins you bought:

• $x_1$ = number of 1L tins
• $x_2$ = number of 2L tins
• $x_3$ = number of 3L tins
• $x_4$ = number of 5L tins

According to the problem, we have two key pieces of information:

1. The total number of tins is 13: $x_1 + x_2 + x_3 + x_4 = 13$

2. The total volume of paint is 29 liters: $1x_1 + 2x_2 + 3x_3 + 5x_4 = 29$

Additionally, we know that at least one tin of each size was bought, so: $x_1 \geq 1, \quad x_2 \geq 1, \quad x_3 \geq 1, \quad x_4 \geq 1$

Now, let's solve this system of equations. Substituting the values to simplify:

Let $y_1 = x_1 - 1$, $y_2 = x_2 - 1$, $y_3 = x_3 - 1$, and $y_4 = x_4 - 1$. This substitution accounts for the fact that there is at least one tin of each size:

$(y_1 + 1) + (y_2 + 1) + (y_3 + 1) + (y_4 + 1) = 13$ Simplifies to: $y_1 + y_2 + y_3 + y_4 = 9$

Similarly, for the volume equation: $1(y_1 + 1) + 2(y_2 + 1) + 3(y_3 + 1) + 5(y_4 + 1) = 29$ Simplifies to: $y_1 + 2y_2 + 3y_3 + 5y_4 = 19$

Now, we solve for $y_1$, ( y_