Given that the incorrect product of reversing the digits of one of the two-digit numbers is 551, we know the correct product must be even. Let's solve the problem step by step.

Step 1: Identify possible two-digit numbers

Let's define the two-digit number as $10a + b$ (where $a$ is the tens digit and $b$ is the units digit). The reversed number will be $10b + a$.

Step 2: Set up the equation for the incorrect product

Given the incorrect product is 551, we can write the following equation for the incorrect multiplication: $(10b + a) \cdot y = 551$

Step 3: Factor 551

551 is an odd number, so we need to find its factors to determine possible values for $10b + a$ and $y$: $551 = 19 \times 29$

Thus, the possible pairs $(10b + a, y)$ are:

1. $(19, 29)$
2. $(29, 19)$

Step 4: Determine the correct product

Case 1: $10b + a = 19$ and $y = 29$

For the original number, we need: $10a + b = x$

If $10b + a = 19$, this implies $b = 1$ and $a = 9$. So, $10a + b = 10 \cdot 9 + 1 = 91$.

The correct product is: $91 \times 29 = 2639$

2639 is not even, so this case is not possible.

Case 2: $10b + a = 29$ and $y = 19$

For the original number, we need: $10a + b = x$

If $10b + a = 29$, this implies $b = 2$ and $a = 9$. So, $10a + b = 10 \cdot 9 + 2 = 92$.

The correct product is: $92 \times 19$

Let's calculate: $92 \times 19 = 92 \times (20 - 1) = 92 \times 20 - 92 = 1840 - 92 = 1748$

1748 is even, so this case is correct.

Conclusion

The correct product of the multiplication is: $\boxed{1748}$

Would you like further details on the reasoning or calculation process? Here are some additional questions to expand your understanding:

1. What are the possible values of $a$ and $b$ if the two-digit number has a particular sum?
2. How would the problem change if both numbers $x$ and $y$ were multi-digit numbers?
3. Can you provide an example where $a$ and $b$ satisfy the given conditions, and compute the exact product?
4. What other number properties can be inferred from the conditions given in the problem?
5. What happens if the mistake was made in reversing the digits of $y$ instead of $x$?
6. How would the solution change if both $x$ and $y$ were prime numbers?
7. If $y$ were an even number, how would it affect the determination of $x$ to ensure an even product?
8. Could you generalize this problem for three-digit numbers with similar reversal conditions?

Tip: When dealing with problems involving digit reversal, always consider the properties of the digits (odd or even) to determine the correct multiplication outcome.