Answer:
![A=2044](https://tex.z-dn.net/?f=A%3D2044)
Step-by-step explanation:
By definition, consecutive numbers follow each other when we count up (e.g. 1, 2, 3).
Let's consider our conditions:
- A, B, and C are consecutive whole numbers greater than 2,000
- A is a multiple of 4
- B is a multiple of 5
- C is a multiple of 6
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:
![2000=2\mod6](https://tex.z-dn.net/?f=2000%3D2%5Cmod6)
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:
![C=1998+18=2016](https://tex.z-dn.net/?f=C%3D1998%2B18%3D2016)
Following the conditions given in the problem, the following must be true:
![A\in \mathbb{Z},\\B\in \mathbb{Z},\\C\in \mathbb{Z},\\A+1=B=C-1,\\A=0\mod 4,\\B=0\mod 5,\\C=0\mod 6,](https://tex.z-dn.net/?f=A%5Cin%20%5Cmathbb%7BZ%7D%2C%5C%5CB%5Cin%20%5Cmathbb%7BZ%7D%2C%5C%5CC%5Cin%20%5Cmathbb%7BZ%7D%2C%5C%5CA%2B1%3DB%3DC-1%2C%5C%5CA%3D0%5Cmod%204%2C%5C%5CB%3D0%5Cmod%205%2C%5C%5CC%3D0%5Cmod%206%2C)
For
, we have
and
:
![A\in \mathbb{Z},\checkmark\\B\in \mathbb{Z},\checkmark\\C\in \mathbb{Z},\checkmark\\A+1=B=C-1,\checkmark\\A=2014\neq 0\mod 6, \times\\B=2015=0\mod 5,\checkmark\\C=2016=0\mod 6\checkmark\\](https://tex.z-dn.net/?f=A%5Cin%20%5Cmathbb%7BZ%7D%2C%5Ccheckmark%5C%5CB%5Cin%20%5Cmathbb%7BZ%7D%2C%5Ccheckmark%5C%5CC%5Cin%20%5Cmathbb%7BZ%7D%2C%5Ccheckmark%5C%5CA%2B1%3DB%3DC-1%2C%5Ccheckmark%5C%5CA%3D2014%5Cneq%200%5Cmod%206%2C%20%5Ctimes%5C%5CB%3D2015%3D0%5Cmod%205%2C%5Ccheckmark%5C%5CC%3D2016%3D0%5Cmod%206%5Ccheckmark%5C%5C)
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:
![A\in \mathbb{Z},\checkmark\\B\in \mathbb{Z},\checkmark\\C\in \mathbb{Z},\checkmark\\A+1=B=C-1,\checkmark\\A=2044=0\mod 4,\checkmark\\B=2045=0\mod 5,\checkmark\\C=2046=0\mod 6\checkmark](https://tex.z-dn.net/?f=A%5Cin%20%5Cmathbb%7BZ%7D%2C%5Ccheckmark%5C%5CB%5Cin%20%5Cmathbb%7BZ%7D%2C%5Ccheckmark%5C%5CC%5Cin%20%5Cmathbb%7BZ%7D%2C%5Ccheckmark%5C%5CA%2B1%3DB%3DC-1%2C%5Ccheckmark%5C%5CA%3D2044%3D0%5Cmod%204%2C%5Ccheckmark%5C%5CB%3D2045%3D0%5Cmod%205%2C%5Ccheckmark%5C%5CC%3D2046%3D0%5Cmod%206%5Ccheckmark)
All conditions are met and therefore our answer is ![\boxed{2,044}](https://tex.z-dn.net/?f=%5Cboxed%7B2%2C044%7D)