The following table is part of mutliplicatin table and some of the digits in the product are replaced with question marks. For example, \(x \cdot y = 7?2\). What is the sum of the four product numbers in this table?

  y y + 1
x 7?2 ??9
x+1 ?2? 8??
Solutions

According to the table, we have four equations.

\[\begin{align} x \cdot y = 7?2 \\ x \cdot (y + 1) = ??9 \\ (x+1) \cdot y = ?2? \\ (x+1) \cdot (y + 1) = 8?? \end{align}\]

Looking at the first two equations, we can see \(7?2 + x = ??9\).

\[\begin{align} x \cdot y = 7?2 \\ x \cdot (y + 1) = x \cdot y + x = 7?2 + x = ??9 \\ \end{align}\]

Therefore, the unit digit of \(x\) must be 7.

Since \(x \cdot y = 7?2\), the unit digit of \(y\) must be 6. Also, the unit digit of \(x + 1\) must be 8, and that of \(y + 1\) must be 7.

With the information about unit digits, we can find out the unit digit of all the product numbers.

  y y + 1
x 7?2 ??9
x+1 ?28 8?6

Now we focus on the following equation because there is only one digit missing and we have clues about the hundred digit.

\[(x+1) \cdot y = ?28\]

The question mark in this equation is either 7 or 8. We try 8 here (you can try 7, but it does not work). If the question mark is 8, we have

\[(x+1) \cdot y = 828.\]

Now we need to find two integers. One of them has 8 at the unit place, the other has 6 at the unit place, and their product is 828.

Let us find all prime factors of 828.

\[828 = 2 \times 2 \times 3 \times 3 \times 23.\]

One pair you may find is \(x + 1 = 138\) and \(y = 6\). However, they do not work for ALL equations.

Another pair is \(x + 1 = 18\) and \(y = 46\). Voila! They work for all equations. So, \(x = 17\) and \(y = 46\).

  46 47
17 782 799
18 828 846

Finally, the answer is 782 + 799 +828 + 846 = 3255.