2. Writer Printko prints the digit 1 with a width of $2 \mathrm{~mm}$, and the remaining digits are $6 \mathrm{~mm}$ wide. The space between each two adjacent digits is $1 \mathrm{~mm}$. Calculate the width of the product of the numbers 2367 and 357.
2. The sought product is $2367 \cdot 357=845019$. 2 POINTS

In the product, there are 5 digits with a width of $6 \mathrm{~mm}$ and 1 digit with a width of $2 \mathrm{~mm}$, so their total width is

$5 \cdot 6 + 1 \cdot 2 = 32 \mathrm{~mm}$. 2 POINTS

Between 6 digits, there are 5 spaces whose total width is $5 \cdot 1 = 5 \mathrm{~mm}$. 1 POINT

The total width of the product 845019 is $32 + 5 = 37 \mathrm{~mm}$. 1 POINT

TOTAL 6 POINTS