https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Mathwizard38025&feedformat=atomAoPS Wiki - User contributions [en]2024-03-29T13:19:29ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_8&diff=1274812000 AMC 8 Problems/Problem 82020-07-05T01:50:09Z<p>Mathwizard38025: /* Problem */</p>
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<div>-<br />
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==Problem==<br />
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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<br />
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==Solution==<br />
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The numbers on one die total <math>1+2+3+4+5+6 = 21</math>, so the numbers<br />
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>.<br />
This leaves <math>63 - 22 = \boxed{\text{(D) 41}}</math> not seen.<br />
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==See Also==<br />
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{{AMC8 box|year=2000|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathwizard38025https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_8&diff=1274802000 AMC 8 Problems/Problem 82020-07-05T01:49:19Z<p>Mathwizard38025: /* Solution */</p>
<hr />
<div>-<br />
<br />
==Problem==<br />
<br />
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<br />
<br />
[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]<br />
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==Solution==<br />
<br />
The numbers on one die total <math>1+2+3+4+5+6 = 21</math>, so the numbers<br />
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>.<br />
This leaves <math>63 - 22 = \boxed{\text{(D) 41}}</math> not seen.<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathwizard38025https://artofproblemsolving.com/wiki/index.php?title=2000_AMC_8_Problems/Problem_8&diff=1274792000 AMC 8 Problems/Problem 82020-07-05T01:47:13Z<p>Mathwizard38025: /* Problem */</p>
<hr />
<div>-<br />
<br />
==Solution==<br />
<br />
The numbers on one die total <math>1+2+3+4+5+6 = 21</math>, so the numbers<br />
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>.<br />
This leaves <math>63 - 22 = \boxed{\text{(D) 41}}</math> not seen.<br />
<br />
==See Also==<br />
<br />
{{AMC8 box|year=2000|num-b=7|num-a=9}}<br />
{{MAA Notice}}</div>Mathwizard38025