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

Let $d_i$ be the number on the $i$ th die. There is a symmetry where we can replace each die's result with $d_i' = 7-d_i$ . Note that applying the symmetry twice we get back to where we started since $7-(7-d_i)=d_i$ . Under this symmetry the sum $S=\sum_{i=1}^n d_i$ is replaced by $S' = \sum_{i=1}^n 7-d_i = 7n - S$ . As a result of this symmetry the probabilities of obtaing the sum $S$ and the sum $S'$ are equal because any combination of $d_i$ which sum to $S$ can be replaced with $d_i'$ to get the sum $S'$ , and conversely. In other words, there is a one-to-one mapping between the combinations of dice which sum to $S$ and the combinations which sum to $S'$ .

The smallest non-zero probability of obtaining $S=1994$ occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in $S=1994$ being impossible. That happens when $n = \left\lceil \frac{1994}{6} \right\rceil = 333$ . By the symmetry mentioned above, the probability of rolling $S=1994$ is the same as the probability of rolling $S' = 333\cdot 7 - 1994 = 337$ . The answer is $\textbf{(C)}$ .

1994 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 29	Followed by Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1994 AHSME Problems/Problem 30

Problem

Solution

See Also