Difference between revisions of "1994 AHSME Problems/Problem 30"
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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>. | 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>. | ||
− | + | When <math>S=1994</math> the smallest number <math>S'=7n-1994</math> corresponds to the smallest number <math>n</math>. Thus we want to find the smallest <math>n</math> which gives non-zero probability of obtaining <math>S=1994</math>. This 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. Clearly <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>. | |
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==See Also== | ==See Also== |
Revision as of 14:19, 28 May 2021
Problem
When 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 . The smallest possible value of is
Solution
Let be the number on the th die. There is a symmetry where we can replace each die's result with . Note that applying the symmetry twice we get back to where we started since . Under this symmetry the sum is replaced by . As a result of this symmetry the probabilities of obtaing the sum and the sum are equal because any combination of which sum to can be replaced with to get the sum , and conversely. In other words, there is a one-to-one mapping between the combinations of dice which sum to and the combinations which sum to .
When the smallest number corresponds to the smallest number . Thus we want to find the smallest which gives non-zero probability of obtaining . This occurs when there are just enough dice for this sum to be possible, and any fewer dice would result in being impossible. Clearly . By the symmetry mentioned above, the probability of rolling is the same as the probability of rolling . The answer is .
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 29 |
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