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Difference between revisions of "2000 AMC 8 Problems/Problem 8"
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The numbers on one die total <math>1+2+3+4+5+6 = 21</math>, so the numbers
 
The numbers on one die total <math>1+2+3+4+5+6 = 21</math>, so the numbers
 
on the three dice total <math>63</math>. Numbers <math>1, 1, 2, 3, 4, 5, 6</math> are visible, and these total <math>22</math>.
 
on the three dice total <math>63</math>. Numbers <math>1, 1, 2, 3, 4, 5, 6</math> are visible, and these total <math>22</math>.
−
This leaves <math>63 - 22 = \boxed{{\text{(B) 41}}</math> not seen.
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This leaves <math>63 - 22 = \boxed{\text{(D) 41}}</math> not seen.
  
 
==See Also==
 
==See Also==
  
 
{{AMC8 box|year=2000|num-b=7|num-a=9}}
 
{{AMC8 box|year=2000|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 06:57, 20 December 2024

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

[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]

$\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 31 \qquad \text{(D)}\ 41 \qquad \text{(E)}\ 53$

Solution

The numbers on one die total $1+2+3+4+5+6 = 21$, so the numbers on the three dice total $63$. Numbers $1, 1, 2, 3, 4, 5, 6$ are visible, and these total $22$. This leaves $63 - 22 = \boxed{\text{(D) 41}}$ 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. AMC logo.png

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