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

Answer:

$A=2044$

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:

$2000=2\mod6$

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:

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

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

$A\in \mathbb{Z},\checkmark\\B\in \mathbb{Z},\checkmark\\C\in \mathbb{Z},\checkmark\\A+1=B=C-1,\checkmark\\A=2014\neq 0\mod 6, \times\\B=2015=0\mod 5,\checkmark\\C=2016=0\mod 6\checkmark$

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:

$A\in \mathbb{Z},\checkmark\\B\in \mathbb{Z},\checkmark\\C\in \mathbb{Z},\checkmark\\A+1=B=C-1,\checkmark\\A=2044=0\mod 4,\checkmark\\B=2045=0\mod 5,\checkmark\\C=2046=0\mod 6\checkmark$

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

Answer 2

Answer:

$, B, and C are consecutive whole numbers greater than 2,000A is a multiple of 4B is a multiple of 5C is a multiple of 6Since B is a multiple of 5, the ones digit of B must be either 0 or 5. However, notice that the number before it, A, needs to be a multiple of 4. The ones digit of a number preceding a ones digit of 0 is 9. There are no multiples of 4 that have a ones digit of 9 and therefore the ones digit of B must be 5.Because of this, we've identified that the ones digit of A, B, and C must be 4, 5, and 6 respectively.We can continue making progress by trying to identify the smallest possible whole number greater than 2,000 with a units digit of 6 that is divisible by 6. Notice that:2000=2\mod62000=2mod6Therefore, 2000-2=19982000−2=1998 must be divisible by 6. To achieve a units digit of 6, we need to add a number with a units digit of 8 to 1,998 (since 8+8 has a units digit of 6).The smallest multiple of 6 that has a units digit of 8 is 18. Check to see if this works:C=1998+18=2016C=1998+18=2016$