Ewan writes out a sequence where he counts by 11 s starting at 3 . The resulting sequence is $3,14,25,36, \ldots$ A number that will appear in Ewan's sequence is
(A) 113
(B) 111
(C) 112
(D) 110
(E) 114
Ewan's sequence starts with 3 and each following number is 11 larger than the previous number. Since every number in the sequence is some number of 11 s more than 3 , this means that each number in the sequence is 3 more than a multiple of 11. Furthermore, every such positive integer is in Ewan's sequence.

Since $110=11 \times 10$ is a multiple of 11 , then $113=110+3$ is 3 more than a multiple of 11 , and so is in Ewan's sequence.

Alternatively, we could write Ewan's sequence out until we get into the correct range:

$$
3,14,25,36,47,58,69,80,91,102,113,124, \ldots
$$