HMMT February, 2023

Combinatorics Difficulty 5.4 AIME, harder Prove it United States

Problem:

Let $A$ , $E$ , $H$ , $L$ , $T$ , and $V$ be chosen independently and at random from the set $\left\{0, \frac{1}{2}, 1\right\}$ . Compute the probability that $\lfloor T \cdot H \cdot E\rfloor = L \cdot A \cdot V \cdot A$ .

Solution

Solution:

There are $3^{3} - 2^{3} = 19$ ways to choose $L$ , $A$ , and $V$ such that $L \cdot A \cdot V \cdot A = 0$ , since at least one of $\{L, A, V\}$ must be $0$ , and $3^{3} - 1 = 26$ ways to choose $T$ , $H$ , and $E$ such that $\lfloor T \cdot H \cdot E\rfloor = 0$ , since at least one of $\{T, H, E\}$ must not be $1$ , for a total of $19 \cdot 26 = 494$ ways. There is only one way to make $\lfloor T \cdot H \cdot E\rfloor = L \cdot A \cdot V \cdot A = 1$ , namely setting every variable equal to $1$ , so there are $495$ total ways that work out of a possible $3^{6} = 729$ , for a probability of $\frac{55}{81}$ .

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