The product of three consecutive numbers is in the form of $87 X X X X X 8$. Determine the 5 missing digits. (The $X$s do not necessarily represent the same digits.)

This was the original problem statement, the following is the corrected version that appeared in January 1981:

The product of three consecutive even numbers is in the form of $87 X X X X X 8$. Determine the 5 missing digits. (The $X$s do not necessarily represent the same digits.)
The original problem statement appeared with an error, the word "even" was missing. Thus, the problem has no solution. The last digit of the sought number is the number in the units place of the product of the last digits of three consecutive numbers. We can quickly convince ourselves that the product of three consecutive integers cannot end in 8. If the last digits include 5 or 0, then the product ends in 0. The remaining four possibilities are:

$$
1 \cdot 2 \cdot 3=6 ; \quad 2 \cdot 3 \cdot 4=24 ; \quad 6 \cdot 7 \cdot 8=336 ; \quad 7 \cdot 8 \cdot 9=504
$$

However, the product of three consecutive even numbers can be of the form $87 X X X X X 8$. None of the three numbers can end in 0. Since $2 \cdot 4 \cdot 6=48 ; 4 \cdot 6 \cdot 8=192$; thus, the last digits of the three numbers are 2, 4, and 6, respectively. The largest of these numbers is greater than 440, since $440^{3}=85184000$. The smallest of these numbers is less than 450, since $450^{3}=91125000$. The three conditions are satisfied by only one set of numbers: 442; 444; 446. Their product, 87526608, is of the desired form, so the missing 5 digits are: $5 ; 2 ; 6 ; 6$ and 0. (Sz. Cs.)