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Difference between revisions of "2024 SSMO Accuracy Round Problems/Problem 6"

(Created page with "==Problem== Three six-sided dice are rolled. Then, the product of the three numbers on the top faces in calculated. If the probability of getting the product that is not a pe...")
 
 
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==Solution==
 
==Solution==
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For simplicity, we will count the number of ways of getting a product that is a perfect square. We will do casework on the number of perfect squares rolled. Firstly, there are <math>2^3 = 8</math> ways to roll 3 perfect squares. Next, if two perfects squares rolled and the other isn't a perfect square, it is impossible for the final product to be a perfect square. If 1 perfect square is rolled, then the other two numbers rolled must be equal. There are <math>3</math> ways to permute the rolls, <math>2</math> perfect squares, and <math>4</math> non-perfect squares, giving <math>3\cdot2\cdot4 = 24</math> possibilities. Finally, if no perfect squares are rolled, we must have <math>236,</math> giving another 6 possibilities. In conclusion, the answer is <cmath>1-\frac{8+24+6}{6^3} = 1-\frac{19}{108} = \frac{89}{108}\implies 89+108 = \boxed{197}.</cmath>
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~SMO_Team

Latest revision as of 14:32, 10 September 2025

Problem

Three six-sided dice are rolled. Then, the product of the three numbers on the top faces in calculated. If the probability of getting the product that is not a perfect square can be expressed as $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers, what is $m+n$?

Solution

For simplicity, we will count the number of ways of getting a product that is a perfect square. We will do casework on the number of perfect squares rolled. Firstly, there are $2^3 = 8$ ways to roll 3 perfect squares. Next, if two perfects squares rolled and the other isn't a perfect square, it is impossible for the final product to be a perfect square. If 1 perfect square is rolled, then the other two numbers rolled must be equal. There are $3$ ways to permute the rolls, $2$ perfect squares, and $4$ non-perfect squares, giving $3\cdot2\cdot4 = 24$ possibilities. Finally, if no perfect squares are rolled, we must have $236,$ giving another 6 possibilities. In conclusion, the answer is \[1-\frac{8+24+6}{6^3} = 1-\frac{19}{108} = \frac{89}{108}\implies 89+108 = \boxed{197}.\]

~SMO_Team

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