1997 PMWC Problems/Problem I12

Revision as of 00:50, 8 August 2009 by Ihatepie (talk | contribs) (Solution)

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

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


Solution 2

Let $a$, $b$, $c$, and $d$ be the numbers on the top of dies one and two and the numbers on the bottom of dies one and two respectively.

Therefore, you are trying to find the sum of $ab+ad+cb+cd$. This factored is $(a+c)(b+d)$. Since the sum of opposite faces is $7$, the answer is $7*7$ or $49$

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
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