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Difference between revisions of "1994 AHSME Problems/Problem 30"

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<math> \textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ 341 </math>
 
<math> \textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ 341 </math>
 
==Solution==
 
==Solution==
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Since the sum of dice is <math>1994</math>, there are at least <math>333</math> dices. Now consider the following numbers <math>\underbrace { 6+6+ \dots + 6 }_{ 332 } + 2</math> and <math>\underbrace{ 1+1+ \dots + 1 }_{ 332 } + 5</math>. They both have the same probability of occurring because <math>6</math> negated by switching it to <math>1</math> and <math>2</math> is negated by switching it to <math>5</math>. So we have obtained an upper bound over <math>S</math> which is <math>1 \cdot 332 + 5 = 337</math>. Only smaller numbers to consider are <math>1 \cdot 332 + 4</math>, <math>1 \cdot 332 + 3</math>, <math>1 \cdot 332 + 2</math>, <math>1 \cdot 332 + 1</math>. As it turns out, none of these work.
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Given that there are <math>n</math> dice, we know that the chance of rolling a sum of <math>1994</math> is the same as that of rolling a sum of <math>7n - 1994</math>. This is because there exists a bijection between the set <math>A</math> of dice rolls that sum to <math>1994</math> and the set <math>B</math> of dice rolls that sum to <math>7n - 1994</math>. In other words, for every ordered n-tuple <math>(a_1, a_2, \dots, a_n) \in A</math> such that <math>a_i \in \{1,2,3,4,5,6\}</math> for all valid <math>i</math> and that <math>\sum^{1994}_{i=0} a_i = 1994</math>, there is a unique ordered n-tuple <math>(b_1, b_2, \dots, b_n) \in B</math> where <math>b_i = 7 - a_i</math> for all valid <math>i</math>, whose sum is indeed <math>7n - 1994</math>.
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As such, minimizing <math>S</math> is the same as minimizing <math>n</math>. The minimum value of <math>n</math> is <math>\left\lceil \frac{1994}{6} \right\rceil = 333</math>. Hence, <math>S = 7n - 1994 = 337</math> <math>\textbf{(C)}</math>.

Revision as of 15:19, 30 December 2018

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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