# 1983 AHSME Problems/Problem 16

## Problem

Let $x = .123456789101112....998999$, where the digits are obtained by writing the integers $1$ through $999$ in order. The $1983$rd digit to the right of the decimal point is $\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 8$

## Solution

We consider the first $1983$ digits, letting the $1983$rd digit be $z$. We can break the string of digits into three segments: let $A$ denote $123456789$ (the $1$-digit numbers), let $B$ denote $1011...9899$ (the $2$-digit numbers), and let $C$ denote $100101...z$ (the $3$-digit numbers). Clearly there are $9$ digits in $A$; in $B$, there are $99-10+1 = 90$ numbers, so $90 \cdot 2 = 180$ digits. This leaves $1983 - 9 - 180 = 1794$ digits in $C$. Notice that $1794 = 3 \cdot 598$ with no remainder, so $C$ consists of precisely the first $598$ $3$-digit numbers. Since the first $3$-digit number is $100$, the $598$th is $100 + 598 - 1 = 697$, so as $z$ is the last digit, the answer is $\boxed{\textbf{(D)}\ 7}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 