2000 AMC 8 Problems/Problem 8
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[hide]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
![[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]](http://latex.artofproblemsolving.com/e/6/e/e6edba55d99924ea2b25e28f44f68200e0b47bed.png)
Solution
Solution 1
The numbers on one die total 
, so the numbers
on the three dice total 
. Numbers 
are visible, and these total 
.
This leaves $63 - 22 = \boxed{{\text{(B) 41}} not seen.
