13. Joe and Blair take turns counting from 1, each person counting one more number than the last number counted in the previous round. If Joe starts with 1, Blair then counts 1, 2, Joe then counts 1, 2, 3, and so on, then the 53rd number counted is ( ).
(A) 2
(B) 3
(C) 5
(D) 6
(E) 8
13. E.

Write the sequence $1,1,2,1,2,3, \cdots$ as
(1) , $(1,2),(1,2,3), \cdots$

Then the last number of the $k$-th group is $k$.
Thus, $a_{1+2+\cdots+k}=a_{\frac{k(k+1)}{2}}=k$.
Also, $\frac{9 \times 10}{2}=45<53<55=\frac{10 \times 11}{2}$, so the 53rd number is the 8th number in the 10th group, which is 8.