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

(New page: 49)
 
(sol)
Line 1: Line 1:
49
+
== 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 <math>x,y</math> be the two numbers at the top of the two dice. Then
 +
 
 +
<cmath>xy + x(7-y) + (7-x)y + (7-x)(7-y) = (x + (7-x))(y + (7 - y)) = 49</cmath>
 +
 
 +
== See also ==
 +
{{PMWC box|year=1997|num-b=I11|num-a=I13}}
 +
 
 +
[[Category:Introductory Combinatorics Problems]]

Revision as of 18:07, 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