Difference between revisions of "2008 AMC 8 Problems/Problem 2"
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==Solution== | ==Solution== | ||
We can derive that <math>C=8</math>, <math>L=6</math>, <math>U=7</math>, and <math>E=1</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 8671}</math> | We can derive that <math>C=8</math>, <math>L=6</math>, <math>U=7</math>, and <math>E=1</math>. Therefore, the answer is <math>\boxed{\textbf{(A)}\ 8671}</math> | ||
| + | ~edited by Owencheng | ||
| + | |||
| + | ==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{\textbf{(A)}\ 8671}</math>. ~solution by SuperVince1 | ||
==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}} | ||
Latest revision as of 10:17, 25 May 2024
Contents
Problem
The ten-letter code 
represents the ten digits 
, in order. What 4-digit number is represented by the code word 
?
Solution
We can derive that 
, 
, 
, and 
. Therefore, the answer is 
~edited by Owencheng
Solution 2
We can easily see that , which makes only 
and 
possible. Since the only difference between answer choices 
and 
is the last digit and that the last digit in the code word "CLUE" is 
, we can just find that and that the answer is . ~solution by SuperVince1
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. 
