Three spinners are shown. The spinners are used to determine the hundreds, tens and ones digits of a threedigit number. How many possible three-digit numbers that can be formed in this way are divisible by 6 ?
(A) 11
(B) 16
(C) 22
(D) 12
(E) 9

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-017.jpg?height=176&width=455&top_left_y=888&top_left_x=1277)
A number is divisible by 6 if it is divisible by both 2 and 3 .

To be divisible by 2 , the three-digit number that is formed must be even and so the ones digit must be 0 or 2 .

To be divisible by 3 , the sum of the digits of the number must be a multiple of 3 .

Consider the possible tens and hundreds digits when the ones digit is 0 .

In this case, the sum of the tens and hundreds digits must be a multiple of 3 (since the ones digit does not add anything to the sum of the digits).

We determine the possible sums of the tens and hundreds digits in the table below.

The sums which are a multiple of 3 are circled.

The Tens Digit

| $100 \mathrm{~s}$ | 5 | 6 | 7 | 8 |
| :---: | :---: | :---: | :---: | :---: |
| 1 | (6) | 7 | 8 | (9) |
| 2 | 7 | 8 | (9) | 10 |
| 3 | 8 | (9) | 10 | 11 |
| 4 | (9) | 10 | 11 | $(12$ |

When the ones digit is 0 , the possible three-digit numbers are: $150,180,270,360,450$, and 480 . Consider the possible tens and hundreds digits when the ones digit is 2 .

In this case, the sum of the tens and hundreds digits must be 2 less than a multiple of 3 (since the ones digit adds 2 to the sum of the digits).

When the ones digit is 2, the possible three-digit numbers are: $162,252,282,372$, and 462 .

The number of three-digit numbers that can be formed that are divisible by 6 is 11 .

ANSWER: (A)