Steve begins at 7 and counts forward by 3 , obtaining the list $7,10,13$, and so on. Dave begins at 2011 and counts backwards by 5, obtaining the list 2011,2006,2001, and so on. Which of the following numbers appear in each of their lists?
(A) 1009
(B) 1006
(C) 1003
(D) 1001
(E) 1011
Steve counts forward by 3 beginning at 7 .

That is, the numbers that Steve counts are each 7 more than some multiple of 3.

We can check the given answers to see if they satisfy this requirement by subtracting 7 from each of them and then determining if the resulting number is divisible by 3 .

We summarize the results in the table below.

| Answers | Result after subtracting 7 | Divisible by 3? |
| :---: | :---: | :---: |
| 1009 | 1002 | Yes |
| 1006 | 999 | Yes |
| 1003 | 996 | Yes |
| 1001 | 994 | No |
| 1011 | 1004 | No |

Of the possible answers, Steve only counted 1009, 1006 and 1003.

Dave counts backward by 5 beginning at 2011.

That is, the numbers that Dave counts are each some multiple of 5 less than 2011.

We can check the given answers to see if they satisfy this requirement by subtracting each of them from 2011 and then determining if the resulting number is divisible by 5 .

We summarize the results in the table below.

| Answers | Result after being subtracted from 2011 | Divisible by $5 ?$ |
| :---: | :---: | :---: |
| 1009 | 1002 | No |
| 1006 | 1005 | Yes |
| 1003 | 1008 | No |
| 1001 | 1010 | Yes |
| 1011 | 1000 | Yes |

Of the possible answers, Dave only counted 1006, 1001 and 1011.

Thus while counting, the only answer that both Steve and Dave will list is 1006.

ANswer: (B)