Calculate $3 \cdot 15+20: 4+1$.

Then add parentheses to the expression so that the result is:

1. the largest possible integer,
2. the smallest possible integer.

(M. Volfová)

Hint. Determine where parentheses can be placed throughout the expression.
Given example without any parentheses yields:

$$
3 \cdot 15+20: 4+1=45+5+1=51 .
$$

Now we will place parentheses and compare the results step by step. We aim to exhaust all possibilities, while ignoring parentheses that are unnecessary, for example:

$$
[3 \cdot(15+20)]: 4+1=3 \cdot(15+20): 4+1
$$

and so on. First, we will list the options with one set of parentheses, followed by those with two sets. It is not possible to place three or more sets of parentheses meaningfully.

$$
\begin{array}{ll}
(3 \cdot 15+20): 4+1=(45+20): 4+1=65: 4+1 & \text { cannot be divided without a remainder } \\
3 \cdot(15+20): 4+1=3 \cdot 35: 4+1=105: 4+1 & \text { cannot be divided without a remainder } \\
3 \cdot(15+20: 4)+1=3 \cdot(15+5)+1=3 \cdot 20+1=61 & \\
3 \cdot(15+20: 4+1)=3 \cdot(15+5+1)=3 \cdot 21=63 & \\
3 \cdot 15+20:(4+1)=45+20: 5=45+4=49 & \\
3 \cdot[(15+20): 4+1]=3 \cdot[35: 4+1] & \\
3 \cdot[15+20:(4+1)]=3 \cdot[15+4]=3 \cdot 19=57 & \\
(3 \cdot 15+20):(4+1)=(45+20): 5=65: 5=13 & \\
3 \cdot(15+20):(4+1)=3 \cdot 35: 5=105: 5=21 &
\end{array}
$$

The largest number we obtained is:

$$
3 \cdot(15+20: 4+1)=63
$$

The smallest number we obtained is:

$$
(3 \cdot 15+20):(4+1)=13 \text {. }
$$

Note. Children will likely place parentheses by trial and error. They may not verify all possibilities; it is likely that they will skip those where division is not exact,
and that, for example, when looking for the largest number, they will skip some cases because the "result would be too small." Such approaches should be recognized.

Recognize as correct also justifications that to obtain the largest whole number, we want the smallest divisor and the largest expression that is multiplied by three; for the smallest whole number, we consider the opposite.

If a student thinks of placing the beginning or end of a parenthesis in the middle of a two-digit number, they can obtain the smallest number as follows:

$$
(3 \cdot 15+2) 0:(4+1)=47 \cdot 0: 5=0 .
$$

Such a solution should also be recognized.