2008 AMC 8 Problem 24

Attempt Problem 24 of the 2008 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 2008 AMC 8 solutions, or check the answer key.

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

24.

Ten tiles numbered 11 through 1010 are turned face down. One tile is turned up at random, and a die is rolled. What is the probability that the product of the numbers on the tile and the die will be a square?
 110 \ \dfrac{1}{10}
 16 \ \dfrac{1}{6}
 1160 \ \dfrac{11}{60}
 15 \ \dfrac{1}{5}
 730 \ \dfrac{7}{30}
Answer: C
Concepts:basic probabilityperfect squarecasework
Small hint:
There are 10⋅610\cdot 6 equally likely tile-die pairs.
Big hint:
For each die roll, count tile numbers that make the product a square.
Solution:
If the rolled number was 1,1, then the tile can be 1,1, 4,4, 9,9, yielding 33 combinations.
If the rolled number was 2,2, then the tile can be 2,2, 8,8, yielding 22 combinations.
If the rolled number was 3,3, then the tile can be 3,3, yielding 11 combination.
If the rolled number was 4,4, then the tile can be 1,1, 4,4, 9,9, yielding 33 combinations.
If the rolled number was 5,5, then the tile can be 5,5, yielding 11 combination.
If the rolled number was 6,6, then the tile can be 6,6, yielding 11 combination.
The total number of combinations is 3+2+1+3+1+1=11.3+2+1+3+1+1=11. There are 6060 combinations each with equal likelihood, so the probability is 1160.\dfrac{11}{60} .
Thus, the answer is C .
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