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 | + | 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}} |
Revision as of 15:36, 15 May 2011
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
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 |