A, B and C, in that order, are three-consecutive whole numbers. Each is greater that 2000. A is a multiple of 4. B is a multiple 5. C is a multiple of 6. What is the smallest possible value of A?

Step-by-step explanation:

Note that denotes that is a whole number.

By definition, consecutive numbers follow each other when we count up (e.g. 1, 2, 3).

Let's consider our conditions:

Since B is a multiple of 5, the ones digit of B must be either 0 or 5. However, notice that the number before it, A, needs to be a multiple of 4. The ones digit of a number preceding a ones digit of 0 is 9. There are no multiples of 4 that have a ones digit of 9 and therefore the ones digit of B must be 5.

Because of this, we've identified that the ones digit of A, B, and C must be 4, 5, and 6 respectively.

We can continue making progress by trying to identify the smallest possible whole number greater than 2,000 with a units digit of 6 that is divisible by 6. Notice that:

Therefore, must be divisible by 6. To achieve a units digit of 6, we need to add a number with a units digit of 8 to 1,998 (since 8+8 has a units digit of 6).

The smallest multiple of 6 that has a units digit of 8 is 18. Check to see if this works:

Following the conditions given in the problem, the following must be true:

For , we have and :

Not all conditions are met, hence this does not work. The next multiple of 6 that has a units digit of 8 is 48. Adding 48 to 1,998, we get .

For , we have and . Checking to see if this works:

All conditions are met and therefore our answer is