Question

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?

Answer 1

`A=2044`

Explanation:

Note that
`x\in\mathbb{W}` denotes that
`x` 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:

• 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`

Therefore,
`2000-2=1998` 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`

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

`A\in \mathbb{W},\\B\in \mathbb{W},\\C\in \mathbb{W},\\A+1=B=C-1,\\A=0\mod 4,\\B=0\mod 5,\\C=0\mod 6,`

For
`C=2016`, we have
`B=2015` and
`A=2014`:

`A\in \mathbb{W},\checkmark\\B\in \mathbb{W},\checkmark\\C\in \mathbb{W},\checkmark\\A+1=B=C-1,\checkmark\\A=2014\\eq 0\mod 6, *\\B=2015=0\mod 5,\checkmark\\C=2016=0\mod 6\checkmark\\`

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
`C=1998+48=2046`.

For
`C=2046`, we have
`B=2045` and
`A=2044`. Checking to see if this works:

`A\in \mathbb{W},\checkmark\\B\in \mathbb{W},\checkmark\\C\in \mathbb{W},\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`

All conditions are met and therefore our answer is
`\boxed{2,044}`