Difference between revisions of "1997 PMWC Problems/Problem I12"

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== Problem ==
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In a die, 1 and 6, 2 and 5, 3 and 4 appear on opposite faces. When 2 dice are thrown, product of numbers appearing on the top and bottom faces of the 2 dice are formed as follows:
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*number on top face of 1st die x number on top face of 2nd die
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*number on top face of 1st die x number on bottom face of 2nd die
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*number on bottom face of 1st die x number on top face of 2nd die
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*number on bottom face of 1st die x number on bottom face of 2nd die
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What is the sum of these 4 products ?
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== Solution ==
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Let <math>x,y</math> be the two numbers at the top of the two dice. Then
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<cmath>xy + x(7-y) + (7-x)y + (7-x)(7-y) = (x + (7-x))(y + (7 - y)) = 49</cmath>
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== See also ==
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{{PMWC box|year=1997|num-b=I11|num-a=I13}}
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[[Category:Introductory Combinatorics Problems]]

Revision as of 19:07, 8 October 2007

Problem

In a die, 1 and 6, 2 and 5, 3 and 4 appear on opposite faces. When 2 dice are thrown, product of numbers appearing on the top and bottom faces of the 2 dice are formed as follows:

  • number on top face of 1st die x number on top face of 2nd die
  • number on top face of 1st die x number on bottom face of 2nd die
  • number on bottom face of 1st die x number on top face of 2nd die
  • number on bottom face of 1st die x number on bottom face of 2nd die

What is the sum of these 4 products ?


Solution

Let $x,y$ be the two numbers at the top of the two dice. Then

\[xy + x(7-y) + (7-x)y + (7-x)(7-y) = (x + (7-x))(y + (7 - y)) = 49\]

See also

1997 PMWC (Problems)
Preceded by
Problem I11
Followed by
Problem I13
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10