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

Revision as of 21:47, 4 July 2020

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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.

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.

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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>
 ==Solution==

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

Revision as of 21:47, 4 July 2020

Solution

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