In each block $1223334444 \cdots 999999999$, there is 1 digit 1, 2 digits 2,3 digits 3 , and so on.

The total number of digits written in each block is $1+2+3+4+5+6+7+8+9=45$.

We note that $1953 \div 45$ gives a quotient of 43 and a remainder of 18 (that is, $1953=45 \times 43+18$ ). Since each block contains 45 digits, then 43 blocks contain $43 \times 45=1935$ digits.

Since $1953-1935=18$, then the $18^{\text {th }}$ digit written in the next block (the $44^{\text {th }}$ block) will be the $1953^{\text {rd }}$ digit written.

Writing out the first 18 digits in a block, we get 122333444455555666 , and so the $1953^{\text {rd }}$ digit written is a 6 .

ANSWER: (C)
The digits from 1 to 9 are written in order so that the digit $n$ is written $n$ times. This forms the block of digits $1223334444 \cdots 999999999$. The block is written 100 times. What is the $1953^{\text {rd }}$ digit written?
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8