Difference between revisions of "1997 AJHSME Problems/Problem 20"
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Thus, the answer is <math>\frac{10}{64} = \frac{5}{32}</math>, and the answer is <math>\boxed{A}</math> | Thus, the answer is <math>\frac{10}{64} = \frac{5}{32}</math>, and the answer is <math>\boxed{A}</math> | ||
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| + | But this may be slightly complicated, so you can also do solution 2 (scroll to bottom). | ||
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| + | == Solution 2 == | ||
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| + | Make a chart with all products. 10 work out of 64. Simplify for 5/32 or <math> \boxed{A} </math>. | ||
== See also == | == See also == | ||
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* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
| − | + | {{MAA Notice}} | |
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Latest revision as of 19:37, 11 January 2025
Contents
Problem
A pair of 8-sided dice have sides numbered 1 through 8. Each side has the same probability (chance) of landing face up. The probability that the product of the two numbers that land face-up exceeds 36 is
Solution
There are 
combinations to examine.
If one die is 
, then even with an 
on the other die, no combinations will work.
If one die is 
, then even with an on the other die, no combinations will work.
If one die is 
, then even with an on the other die, no combinations will work.
If one die is 
, then even with an on the other die, no combinations will work.
If one die is 
, then the other die must be an to have a product over 
. Thus, 
works.
If one die is 
, then the other die must be either 
or to have a product over . Thus, 
and 
both work.
If one die is , then the other die can be 
or to have a product over . Thus, 
, 
, and 
all work.
If one die is , then the other die can be 
or to have a product over . Thus, 
and 
work.
There are a total of 
combinations that work out of a total of 
possibilities.
Thus, the answer is 
, and the answer is 
But this may be slightly complicated, so you can also do solution 2 (scroll to bottom).
Solution 2
Make a chart with all products. 10 work out of 64. Simplify for 5/32 or .
See also
| 1997 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 19 |
Followed by Problem 21 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
