Difference between revisions of "2023 AMC 12B Problems/Problem 23"
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Let <math>n=1</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)</math> 6 possible values.But if the only restriction of the product if that 2x≤n,y≤n,z≤n,we can get (2+1)(1+1)(1+1)=12 possible values,(possible values of real situation)/(possible values of ideal situation)=<math>6/12=0.5</math>. | Let <math>n=1</math>, we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,1,0),(1,0,0),(1,1,0),(2,0,0)</math> 6 possible values.But if the only restriction of the product if that 2x≤n,y≤n,z≤n,we can get (2+1)(1+1)(1+1)=12 possible values,(possible values of real situation)/(possible values of ideal situation)=<math>6/12=0.5</math>. | ||
− | Let <math>n=2</math>,we get | + | Let <math>n=2</math>,we get <math>(x,y,z)=(0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),</math><br /> |
− | <math>(x,y,z)= | + | <math>(2,0,0),(2,0,1),(2,1,0),(2,2,0),(3,0,0),(3,1,0),(4,0,0)</math> 17 possible values. |
− | (0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1), | ||
− | (1,0,0),(1,0,1),(1,1,0),(1,1,1),(1,2,0),<br /> | ||
− | (2,0,0),(2,0,1),(2,1,0),(2,2,0), | ||
− | (3,0,0),(3,1,0),(4,0,0)</math> 17 possible values. | ||
possible values of ideal situation=<math>5*3*3=45</math>,<math>17/45</math>≈<math>0.378</math>. | possible values of ideal situation=<math>5*3*3=45</math>,<math>17/45</math>≈<math>0.378</math>. | ||
Revision as of 08:20, 16 November 2023
Contents
[hide]Problem
When standard six-sided dice are rolled, the product of the numbers rolled can be any of possible values. What is ?
Solution1
The product can be written as
Therefore, we need to find the number of ordered tuples where , , , , are non-negative integers satisfying . We denote this number as .
Denote by the number of ordered tuples where with .
Thus,
Next, we compute .
Denote . Thus, for each given , the range of is from 0 to . Thus, the number of is
Therefore,
By solving , we get .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution2(Informal)
The product can be written as
Let , we get 6 possible values.But if the only restriction of the product if that 2x≤n,y≤n,z≤n,we can get (2+1)(1+1)(1+1)=12 possible values,(possible values of real situation)/(possible values of ideal situation)=.
Let ,we get
17 possible values.
possible values of ideal situation=,≈.
Now we can Predict the trend of the product if n becoming bigger,the quotient of (possible values of real situation)/(possible values of ideal situation) will be smaller and smaller.
Let
,you get possible values of ideal situation=.
,the number=.
,the number=.
,the number= so 6 is not the answer.
,the number=.
,the number=,but ≈ still much smaller than 936.
,the number=,but ≈ still smaller than 936.
,the number=, ≈ is a little bigger 936,but the quotient of (possible values of real situation)/(possible values of ideal situation) is much smaller than 0.378 now, so 10 is probably not the answer,so the answer is .
Check calculation: ,the number=,≈ is much bigger than 936.
~Troublemaker
Video Solution 1 by OmegaLearn
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 | |
All AMC 12 Problems and Solutions |
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