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Difference between revisions of "2000 AMC 8 Problems/Problem 8"

(Created page with '==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…')
 
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<math>\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 31 \qquad \text{(D)}\ 41 \qquad \text{(E)}\ 53</math>
 
<math>\text{(A)}\ 21 \qquad \text{(B)}\ 22 \qquad \text{(C)}\ 31 \qquad \text{(D)}\ 41 \qquad \text{(E)}\ 53</math>
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==Problem==
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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==
 
==Solution==
  
−
===Solution 1===
 
 
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>.
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This leaves $63 - 22 = \boxed{{\text{(B) 41}} not seen.
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This leaves <math>63 - 22 = \boxed{\text{(D) 41}}</math> not seen.
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==See Also==
  
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===Solution 2===
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{{AMC8 box|year=2000|num-b=7|num-a=9}}
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{{MAA Notice}}

Latest revision as of 20:41, 25 September 2021

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]

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

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 $1+2+3+4+5+6 = 21$, 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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