Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1983 AHSME Problems/Problem 16"

(Added a solution)
 
m (Added box at the bottom)
 
Line 13: Line 13:
  
 
We consider the first <math>1983</math> digits, letting the <math>1983</math><sup>rd</sup> digit be <math>z</math>. We can break the string of digits into three segments: let <math>A</math> denote <math>123456789</math> (the <math>1</math>-digit numbers), let <math>B</math> denote <math>1011...9899</math> (the <math>2</math>-digit numbers), and let <math>C</math> denote <math>100101...z</math> (the <math>3</math>-digit numbers). Clearly there are <math>9</math> digits in <math>A</math>; in <math>B</math>, there are <math>99-10+1 = 90</math> numbers, so <math>90 \cdot 2 = 180</math> digits. This leaves <math>1983 - 9 - 180 = 1794</math> digits in <math>C</math>. Notice that <math>1794 = 3 \cdot 598</math> with no remainder, so <math>C</math> consists of precisely the first <math>598</math> <math>3</math>-digit numbers. Since the first <math>3</math>-digit number is <math>100</math>, the <math>598</math><sup>th</sup> is <math>100 + 598 - 1 = 697</math>, so as <math>z</math> is the last digit, the answer is <math>\boxed{\textbf{(D)}\ 7}</math>.
 
We consider the first <math>1983</math> digits, letting the <math>1983</math><sup>rd</sup> digit be <math>z</math>. We can break the string of digits into three segments: let <math>A</math> denote <math>123456789</math> (the <math>1</math>-digit numbers), let <math>B</math> denote <math>1011...9899</math> (the <math>2</math>-digit numbers), and let <math>C</math> denote <math>100101...z</math> (the <math>3</math>-digit numbers). Clearly there are <math>9</math> digits in <math>A</math>; in <math>B</math>, there are <math>99-10+1 = 90</math> numbers, so <math>90 \cdot 2 = 180</math> digits. This leaves <math>1983 - 9 - 180 = 1794</math> digits in <math>C</math>. Notice that <math>1794 = 3 \cdot 598</math> with no remainder, so <math>C</math> consists of precisely the first <math>598</math> <math>3</math>-digit numbers. Since the first <math>3</math>-digit number is <math>100</math>, the <math>598</math><sup>th</sup> is <math>100 + 598 - 1 = 697</math>, so as <math>z</math> is the last digit, the answer is <math>\boxed{\textbf{(D)}\ 7}</math>.
 +
 +
==See Also==
 +
{{AHSME box|year=1983|num-b=15|num-a=17}}
 +
 +
{{MAA Notice}}

Latest revision as of 23:54, 19 February 2019

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 $\boxed{\textbf{(D)}\ 7}$.

See Also

1983 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions


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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1983_AHSME_Problems/Problem_16&oldid=103572"