1997 AMC 8 Problem 20

Attempt Problem 20 of the 1997 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1997 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

20.

A pair of 88-sided dice has sides numbered 11 through 8.8. 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 3636 is
532\dfrac{5}{32}
1164\dfrac{11}{64}
316\dfrac{3}{16}
14\dfrac{1}{4}
12\dfrac{1}{2}
Answer: A
Concepts:dice (probability)casework
Small hint:
For first rolls 55 through 8,8, count which second rolls make the product exceed 3636
Big hint:
Count ordered pairs from two 88-sided dice
Solution:
We can case on the value of the first die. If its value is 1,1, 2,2, 3,3, or 4,4, then it is impossible for the product to be greater than 36.36.
If it is a 5,5, then the other die has to roll an 8,8, otherwise the product is less than 36.36.
If the first die is a 6,6, then the other die must be 77 or 8,8, giving 22 possibilities. If the first die is a 7,7, then the other die must be 6,6, 7,7, or 8,8, giving 33 possibilities.
Finally, if the first roll is an 8,8, the other die can be 5,5, 6,6, 7,7, or 8,8, giving 44 more possibilities.
The total number of working pairs is 1+2+3+4=10, 1 + 2 + 3 + 4 = 10, and the total number of pairs is 82=64.8^2 = 64. The desired probability is then 1064=532. \dfrac{10}{64} = \dfrac{5}{32}.
Thus, A is the correct answer.
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