Difference between revisions of "2000 AMC 8 Problems/Problem 8"
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− | - | + | ==Problem== |
+ | |||
+ | Three dice with faces numbered <math>1</math> through <math>6</math> are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is | ||
+ | |||
+ | <asy> | ||
+ | draw((0,0)--(2,0)--(3,1)--(3,7)--(1,7)--(0,6)--cycle); | ||
+ | draw((3,7)--(2,6)--(0,6)); | ||
+ | draw((3,5)--(2,4)--(0,4)); | ||
+ | draw((3,3)--(2,2)--(0,2)); | ||
+ | draw((2,0)--(2,6)); | ||
+ | |||
+ | dot((1,1)); dot((.5,.5)); dot((1.5,.5)); dot((1.5,1.5)); dot((.5,1.5)); | ||
+ | dot((2.5,1.5)); | ||
+ | dot((.5,2.5)); dot((1.5,2.5)); dot((1.5,3.5)); dot((.5,3.5)); | ||
+ | dot((2.25,2.75)); dot((2.5,3)); dot((2.75,3.25)); dot((2.25,3.75)); dot((2.5,4)); dot((2.75,4.25)); | ||
+ | dot((.5,5.5)); dot((1.5,4.5)); | ||
+ | dot((2.25,4.75)); dot((2.5,5.5)); dot((2.75,6.25)); | ||
+ | dot((1.5,6.5)); | ||
+ | </asy> | ||
+ | |||
+ | <math>\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 31 \qquad \text{(D)}\ 41 \qquad \text{(E)}\ 53</math> | ||
==Problem== | ==Problem== |
Latest revision as of 20:41, 25 September 2021
Contents
Problem
Three dice with faces numbered through are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is
Problem
Three dice with faces numbered 1 through 6 are stacked as shown. Seven of the eighteen faces are visible, leaving eleven faces hidden (back, bottom, between). The total number of dots NOT visible in this view is
Solution
The numbers on one die total , so the numbers on the three dice total . Numbers are visible, and these total . This leaves not seen.
See Also
2000 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.