Difference between revisions of "1997 PMWC Problems/Problem I12"
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Let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> be the numbers on the top of dies one and two and the numbers on the bottom of dies one and two respectively. | Let <math>a</math>, <math>b</math>, <math>c</math>, and <math>d</math> 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 < | + | Therefore, you are trying to find the sum of <cmath>ab+ad+cb+cd</cmath>. This factored is <cmath>(a+c)(b+d)</cmath>. Since the sum of opposite faces is <cmath>7</cmath>, the answer is <cmath>7*7</cmath> or <cmath>49</cmath>. |
== See also == | == See also == |
Revision as of 00:51, 8 August 2009
Contents
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 be the two numbers at the top of the two dice. Then
Solution 2
Let , , , and 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 . This factored is . Since the sum of opposite faces is , the answer is or .
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 |