Difference between revisions of "2023 AMC 12B Problems/Problem 23"
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<math>\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9</math> | <math>\textbf{(A)}~11\qquad\textbf{(B)}~6\qquad\textbf{(C)}~8\qquad\textbf{(D)}~10\qquad\textbf{(E)}~9</math> | ||
− | == | + | ==Solution1== |
The product can be written as | The product can be written as | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
+ | |||
+ | ==Solution2(Informal)== | ||
+ | |||
+ | The product can be written as | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 2^x 3^y 5^z | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | 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 | ||
+ | <math>(x,y,z)=<br /> | ||
+ | (0,0,0),(0,0,1),(0,0,2),(0,1,0),(0,1,1),<br /> | ||
+ | (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),<br /> | ||
+ | (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>. | ||
+ | |||
+ | 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 | ||
+ | <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=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=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=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>. | ||
+ | |||
+ | Check calculation: | ||
+ | <math>n=11</math>,the number=<math>23*12*12=3312</math>,<math>3312*0.378</math>≈<math>1252</math> is much bigger than 936. | ||
+ | |||
+ | ~Troublemaker | ||
==Video Solution 1 by OmegaLearn== | ==Video Solution 1 by OmegaLearn== |
Revision as of 08:17, 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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