One of the strategies in solving problems is to divide them into cases and see if you can eliminate some impossible cases. In this post, I will discuss a popular classic problem which I found the solution to be interesting.

**Problem**

*If A, B, C, and D are distinct non-zero digits, find their values if ABCD × 4 = DBCA.*

If you have not encountered this problem before, you might want to solve it first before reading the solution below.

**Finding the Value of A**

Solution

One of the most obvious hints of the problem is that the value of *A* is only limited to 1 and 2. Why? Because if *A* is greater than 2, 4 × *A *will become a 2-digit number. This will make the product of *ABCD* × 4 a 5-digit number which does not satisfy the problem.

If *A* = 1, then given becomes 1*BCD *× 4 = *DCB*1 .Now, this is impossible because the ones digit of 4 multiplied by *D* should be even. Therefore, *A* must be 2.

**Finding the Value of D**

Now that we found the value of *A*, the next clue is finding the value of *D* since we already have 2 as the ones digit of the product.

*Solution 1*

It is likely that *D* = 8 since it is the product of 2. However, it is also possible that there is a carry over from 4 × *B* if its product is more than 10. If so, the only possible carry over is 1 since if it is greater than 1 it will make the product a 5-digit number.

Now, suppose 1 is the carry over, then *D* = 4 × 2 + 1 = 9. However, this is not possible since the product of D and 4 has 2 as its ones digit. Therefore, the only possible value for D is 8.

*Solution 2*

We look for the value of *D* such that the ones digit of the product of* D* is 2. We calculate all possible cases as shown below.

4(0) = 0

4(1) = 4

4(2) = 8

**4(3) = 12**

4(4) = 16

4(5) = 20

4(6) = 24

4(7) = 28

**4(8) = 32**

4(9) = 36

As we can see, there are two possible cases, *D* = 3, and *D* = 8. Clearly, *D* cannot equal 3 because it should be the product of 4 and** 2 **and a possible carry over which can only be equal to 1. Now, for it to be 3, the carry over must be 5, but the sum 8 + 5 = 13 which will make the product a 5-digit number.

Therefore, D = 8.

**Finding the value of B **

*Solution 1*

From above, *D* = 4 × 2 = 8, means that we did not carry over from the product of *B*. This means that the product of *B* and 4 is less than 10. So, the only possible values for *B* are 1 and 2. But *A* is already equal to 2 and from the given,* A*, *B*,* C* and *D* are distinct digits.

So, B = 1.

*Solution 2*

From above, *D* = 4 × 2 = 8, means that we did not carry over from the product of *B*. This means that the product of *B* and 4 is less than 10. So, the only possible values for B are 1 and 2

We find the values of B by getting the ones place of the result of 4C + 3.

4(0) + 3 = 3

4(1) + 3 = 7

**4(2) + 3 = 11**

4(3) + 3= 15

4(4) + 3= 19

4(5) + 3 = 23

4(6) + 3 = 27

**4(7) + 3 = 31**

4(8) + 3 = 34

4(9) + 3 = 35

The only value less than 3 is 1. So, B = 1.

**Finding the value of C. **

From the two cases above, 11 and 31, C = 2 or C = 7. Since *A*, *B*, *C*, and* D* are distinct digits and since and A = 2, it must be the case that C = 7.

Therefore, the correct answer from the problem above is 2178. So, 2178 × 4 = 8712.

The problem above shows how some problems can be shown by actually eliminating impossible cases.