Yuko likes to play board games. He decides to play a game with his brother Yuri. They must roll 3 dice at the same time and move the squares corresponding to the value of the 3 dice. Yuri started and his dice had the following combination: 2, 4, and 5. On Yuko's turn, he rolled 1, 5, and X. How many squares must Yuko's last die move him so that he is in front of his brother?
Thoughts? Step-by-step reasoning:
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The sum of Yuri's dice gave, 2 + 4 + 5 = 11 squares. For Yuko to get the same sum in the dice, 1 + 5 + X = 11 squares. Solving the equation, X = 11 - 6, X = 5 squares. To stand in front of his brother, Yuko needs to be at least 1 place in front, so his last die will move him, 5 + 1 = 6 squares.
Thus, the answer is 6.
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