Difference between revisions of "2023 AMC 12B Problems/Problem 23"
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | == | + | ==Solution 2 (Cheese)== |
The product can be written as | The product can be written as | ||
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</cmath> | </cmath> | ||
− | + | Letting <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 <math>2x\le n,y\le n,z\le n</math>, we can get <math>(2+1)(1+1)(1+1)=12</math> possible values. We calculate the ratio | |
+ | <cmath>r = \frac{\text{possible values of real situation}}{\text{possible values of ideal situation}} = \frac{6}{12}=0.5.</cmath> | ||
− | + | Letting <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>(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. | + | <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. |
− | + | The number of possibilities in the ideal situation is <math>5*3*3=45</math>, making <math>r = 17/45 \approx 0.378</math>. | |
− | Now we can | + | Now we can predict the trend of <math>r</math>: as <math>n</math> increases, <math>r</math> decreases. |
− | + | Letting | |
− | <math>n=3</math>,you get possible values of ideal situation=<math>7*4*4=112</math>. | + | <math>n=3</math>, you get possible values of ideal situation=<math>7*4*4=112</math>. |
− | <math>n=4</math>,the number=<math>9*5*5=225</math>.<br /> | + | <math>n=4</math>, the number=<math>9*5*5=225</math>.<br /> |
− | <math>n=5</math>,the number=<math>11*6*6=396</math>.<br /> | + | <math>n=5</math>, the number=<math>11*6*6=396</math>.<br /> |
− | <math>n=6</math>,the number=<math>13*7*7=637,637<936</math> so 6 is not the answer.<br /> | + | <math>n=6</math>, the number=<math>13*7*7=637,637<936</math> so 6 is not the answer.<br /> |
− | <math>n=7</math>,the number=<math>15*8*8=960</math>.<br /> | + | <math>n=7</math>, the number=<math>15*8*8=960</math>.<br /> |
− | <math>n=8</math>,the number=<math>17*9*9=1377</math>,but <math>1377*0.378</math>≈<math>521</math> still much smaller than 936.<br /> | + | <math>n=8</math>, the number=<math>17*9*9=1377</math>,but <math>1377*0.378</math>≈<math>521</math> still much smaller than 936.<br /> |
− | <math>n=9</math>,the number=<math>19*10*10=1900</math>,but <math>1900*0.378</math>≈<math>718</math> still smaller than 936.<br /> | + | <math>n=9</math>, the number=<math>19*10*10=1900</math>,but <math>1900*0.378</math>≈<math>718</math> still smaller than 936.<br /> |
− | <math>n=10</math>,the number=<math>21*11*11=2541</math>, <math>2541*0.378</math>≈<math>960</math> 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 <math>\boxed{\textbf{(A) 11}}</math>. | + | <math>n=10</math>, the number=<math>21*11*11=2541</math>, <math>2541*0.378</math>≈<math>960</math> 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 <math>\boxed{\textbf{(A) 11}}</math>. |
Check calculation: | Check calculation: |
Revision as of 12:30, 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)
Solution 2 (Cheese)
The product can be written as
Letting , we get , 6 possible values. But if the only restriction of the product if that , we can get possible values. We calculate the ratio
Letting , we get
, 17 possible values.
The number of possibilities in the ideal situation is , making .
Now we can predict the trend of : as increases, decreases.
Letting
, 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.