Difference between revisions of "1994 AHSME Problems/Problem 30"
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==Solution== | ==Solution== | ||
− | 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 | + | 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>. |
− | When <math>R=1994</math> we seek the smallest number <math>S=7n-1994</math>, | + | 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>. |
==See Also== | ==See Also== |
Latest revision as of 18:43, 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 number with . Note that applying the symmetry twice we get back to where we started since , so this symmetry is its own inverse. Under this symmetry the sum is replaced by . As a result of this symmetry the sum the sum have the same probability because any combination of which sum to can be replaced with which sum to , 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 we seek the smallest number , which clearly happens when is smallest. Therefore 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. Thus and . The answer is .
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
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