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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> dice. 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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Let <math>d_i</math> be the number on the <math>i</math>th die.  There is a symmetry where we can replace each die's number with <math>d_i' = 7-d_i</math>.  Note that applying the symmetry twice we get back to where we started since <math>7-(7-d_i)=d_i</math>, so this symmetry is its own inverse. Under this symmetry the sum <math>R=\sum_{i=1}^n d_i</math> is replaced by <math>S = \sum_{i=1}^n 7-d_i = 7n - R</math>. As a result of this symmetry the sum <math>R</math> the sum <math>S</math> have the same probability because any combination of <math>d_i</math> which sum to <math>R</math> can be replaced with <math>d_i'</math> which sum to <math>S</math>, and conversely.  In other words, there is a one-to-one mapping between the combinations of dice which sum to <math>R</math> and the combinations which sum to <math>S</math>.
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When <math>R=1994</math> we seek the smallest number <math>S=7n-1994</math>, which clearly happens when <math>n</math> is smallest.  Therefore we want to find the smallest <math>n</math> which gives non-zero probability of obtaining <math>R=1994</math>. This occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in <math>R=1994</math> being impossible.  Thus <math> n = \left\lceil \frac{1994}{6} \right\rceil = 333</math> and <math>S = 333\cdot 7 - 1994 = 337</math>.  The answer is <math>\textbf{(C)}</math>.
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==See Also==
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{{AHSME box|year=1994|num-b=29|after= Last Problem}}
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{{MAA Notice}}

Latest revision as of 17:43, 28 May 2021

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 number with $d_i' = 7-d_i$. Note that applying the symmetry twice we get back to where we started since $7-(7-d_i)=d_i$, so this symmetry is its own inverse. Under this symmetry the sum $R=\sum_{i=1}^n d_i$ is replaced by $S = \sum_{i=1}^n 7-d_i = 7n - R$. As a result of this symmetry the sum $R$ the sum $S$ have the same probability because any combination of $d_i$ which sum to $R$ can be replaced with $d_i'$ which sum to $S$, and conversely. In other words, there is a one-to-one mapping between the combinations of dice which sum to $R$ and the combinations which sum to $S$.

When $R=1994$ we seek the smallest number $S=7n-1994$, which clearly happens when $n$ is smallest. Therefore we want to find the smallest $n$ which gives non-zero probability of obtaining $R=1994$. This occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in $R=1994$ being impossible. Thus $n = \left\lceil \frac{1994}{6} \right\rceil = 333$ and $S = 333\cdot 7 - 1994 = 337$. The answer is $\textbf{(C)}$.

See Also

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

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