In the table shown, a sequence starts with 2 in the top left corner. Moving across each row, each box is filled with a number 3 greater than the number to its left. The leftmost number in each row is 3 greater than the greatest in the previous row. When all of the boxes are filled in, the value of $x$ is
(A) 101
(B) 104
(C) 107
(D) 110
(E) 113

| 2 | 5 | 8 | 11 | 14 | 17 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 20 | 23 | 26 | 29 | 32 | 35 |
| 38 |  |  |  |  |  |
|  |  |  |  |  |  |
|  |  |  |  |  |  |
|  |  |  |  |  | $x$ |
Since there are 6 columns and each term in the sequence is 3 greater than the previous term, then the number in each box in the final column is 18 more than the number above it. (Each number in the final column is 6 terms further along in the sequence than the number above it.) Thus, the number in the bottom right box should be $17+5(18)=17+90=107$, since it is 5 rows below the 17 .

ANswer: (C)