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

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==Solution==
 
==Solution==
  
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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 result 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>.  Under this symmetry the sum <math>S=\sum_{i=1}^n d_i</math> is replaced by <math>S' = \sum_{i=1}^n 7-d_i = 7n - S</math>.  As a result of this symmetry the probabilities of obtaing the sum <math>S</math> and the sum <math>S'</math> are equal because any combination of <math>d_i</math> which sum to <math>S</math> can be replaced with <math>d_i'</math> to get the sum <math>S'</math>, and conversely.  In other words, there is a one-to-one mapping between the combinations of dice which sum to <math>S</math> and the combinations which sum to <math>S'</math>.
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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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The smallest non-zero probability of obtaining <math>S=1994</math> occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in <math>S=1994</math> being impossible.  That happens when <math> n = \left\lceil \frac{1994}{6} \right\rceil = 333</math>.  By the symmetry mentioned above, the probability of rolling <math>S=1994</math> is the same as the probability of rolling <math>S' = 333\cdot 7 - 1994 = 337</math>.  The answer is <math>\textbf{(C)}</math>.
 
  
 
==See Also==
 
==See Also==

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)
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Problem 29 		Followed by
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All AHSME Problems and Solutions

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