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 "2008 AMC 8 Problems/Problem 2"

(→‎Solution 2)
(→‎Solution 2)
Line 13: Line 13:
  
 
==Solution 2==
 
==Solution 2==
−
We can easily see that <math>C=8</math>, which makes only <math>(A)</math> <math>8761</math> and <math>(B)</math> <math>8762</math> possible. Since the only difference between answer choices <math>A</math> and <math>B</math> is the last digit and that the last digit in the code word "CLUE" is <math>E</math>, we can just find that <math>E=1</math> and that the answer is <math>/boxed{(A)</math> <math>8761}</math>.
+
We can easily see that <math>C=8</math>, which makes only <math>(A)</math> <math>8761</math> and <math>(B)</math> <math>8762</math> possible. Since the only difference between answer choices <math>A</math> and <math>B</math> is the last digit and that the last digit in the code word "CLUE" is <math>E</math>, we can just find that <math>E=1</math> and that the answer is <math>\boxed{\textbf{(A)}\ 8671}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC8 box|year=2008|num-b=1|num-a=3}}
 
{{AMC8 box|year=2008|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 10:17, 25 May 2024

Problem

The ten-letter code $\text{BEST OF LUCK}$ represents the ten digits $0-9$, in order. What 4-digit number is represented by the code word $\text{CLUE}$?

$\textbf{(A)}\ 8671 \qquad \textbf{(B)}\ 8672 \qquad \textbf{(C)}\ 9781 \qquad \textbf{(D)}\ 9782 \qquad \textbf{(E)}\ 9872$

Solution

We can derive that $C=8$, $L=6$, $U=7$, and $E=1$. Therefore, the answer is $\boxed{\textbf{(A)}\ 8671}$ ~edited by Owencheng

Solution 2

We can easily see that $C=8$, which makes only $(A)$ $8761$ and $(B)$ $8762$ possible. Since the only difference between answer choices $A$ and $B$ is the last digit and that the last digit in the code word "CLUE" is $E$, we can just find that $E=1$ and that the answer is $\boxed{\textbf{(A)}\ 8671}$.

See Also

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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
All AJHSME/AMC 8 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=2008_AMC_8_Problems/Problem_2&oldid=219713"