1994 AHSME Problems/Problem 30

Problem

When $n$ standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of $S$ . The smallest possible value of $S$ is

$\textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ 341$

Solution

Given that there are $n$ dice, we know that the chance of rolling a sum of $1994$ is the same as that of rolling a sum of $7n - 1994$ . This is because there exists a bijection between the set $A$ of dice rolls that sum to $1994$ and the set $B$ of dice rolls that sum to $7n - 1994$ . In other words, for every ordered n-tuple $(a_1, a_2, \dots, a_n) \in A$ such that $a_i \in \{1,2,3,4,5,6\}$ for all valid $i$ and that $\sum^{1994}_{i=0} a_i = 1994$ , there is a unique ordered n-tuple $(b_1, b_2, \dots, b_n) \in B$ where $b_i = 7 - a_i$ for all valid $i$ , whose sum is indeed $7n - 1994$ .

As such, minimizing $S$ is the same as minimizing $n$ . The minimum value of $n$ is $\left\lceil \frac{1994}{6} \right\rceil = 333$ . Hence, $S = 7n - 1994 = 337$ $\textbf{(C)}$ .

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1994 AHSME Problems/Problem 30

Problem

Solution