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Difference between revisions of "1999 AIME Problems/Problem 7"

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== Solution ==
 
== Solution ==
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For each <math>i</math>th switch (designated by <math>x_{i},y_{i},z_{i}</math>), it advances ''itself'' only one time at the <math>i</math>th step; thereafter, only a switch with larger <math>x_{j},y_{j},z_{j}</math> values will advance the <math>i</math>th switch by one step provided <math>d_{i}= 2^{x_{i}}3^{y_{i}}5^{z_{i}}</math> divides <math>d_{j}= 2^{x_{j}}3^{y_{j}}5^{z_{j}}</math>. Let <math>N = 2^{9}3^{9}5^{9}</math> be the max switch label. To find the divisor multiples in the range of <math>d_{i}</math> to <math>N</math>, we consider the exponents of the number <math>\frac{N}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}</math>. In general, the divisor-count of <math>\frac{N}{d}</math> must be a multiple of 4 to ensure that a switch is in position A:
 
For each <math>i</math>th switch (designated by <math>x_{i},y_{i},z_{i}</math>), it advances ''itself'' only one time at the <math>i</math>th step; thereafter, only a switch with larger <math>x_{j},y_{j},z_{j}</math> values will advance the <math>i</math>th switch by one step provided <math>d_{i}= 2^{x_{i}}3^{y_{i}}5^{z_{i}}</math> divides <math>d_{j}= 2^{x_{j}}3^{y_{j}}5^{z_{j}}</math>. Let <math>N = 2^{9}3^{9}5^{9}</math> be the max switch label. To find the divisor multiples in the range of <math>d_{i}</math> to <math>N</math>, we consider the exponents of the number <math>\frac{N}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}</math>. In general, the divisor-count of <math>\frac{N}{d}</math> must be a multiple of 4 to ensure that a switch is in position A:
  
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<math>4n = [(9-x)+1] [(9-y)+1] [(9-z)+1] = (10-x)(10-y)(10-z)</math>, where <math>0 \le x,y,z \le 9.</math>
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<center><math>4n = [(9-x)+1] [(9-y)+1] [(9-z)+1] = (10-x)(10-y)(10-z)</math>, where <math>0 \le x,y,z \le 9.</math></center>
  
 
We consider the cases where the 3 factors above do not contribute multiples of 4.
 
We consider the cases where the 3 factors above do not contribute multiples of 4.
  
 +
*Case of no 2's:
  
−
Case of no 2's:
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:The switches must be <math>(\mathrm{odd})(\mathrm{odd})(\mathrm{odd})</math>. There are <math>5</math> [[odd integer]]s in <math>0</math> to <math>9</math>, so we have <math>5 \times 5 \times 5 = 125</math> ways.
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<math>(odd)(odd)(odd)</math>
 
−
 
 
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There are <math>5</math> [[odd integer]]s in <math>0</math> to <math>9</math>.
 
−
 
 
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We have <math>5 \times 5 \times 5 = 5^{3}= 125</math> ways.
 
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−
 
 
−
 
 
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Case of a single 2:
 
  
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<math>(2\cdot odd)(odd)(odd)</math> or <math>(odd)(2 \cdot odd)(odd)</math> or <math>(odd)(odd)(2 \cdot odd)</math>
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*Case of a single 2:
  
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Since <math>0 \le x,y,z \le 9,</math> the terms <math>2\cdot 1, 2 \cdot 3,</math> and <math>2 \cdot 5</math> are three valid choices for the <math>(2 \cdot odd)</math> factor above.
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:The switches must be one of <math>(2\cdot \mathrm{odd})(\mathrm{odd})(\mathrm{odd})</math> or <math>(\mathrm{odd})(2 \cdot \mathrm{odd})(\mathrm{odd})</math> or <math>(\mathrm{odd})(\mathrm{odd})(2 \cdot \mathrm{odd})</math>.
  
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We have <math>{3\choose{1}} \cdot 3 \cdot 5^{2}= 225</math> ways.  
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:Since <math>0 \le x,y,z \le 9,</math> the terms <math>2\cdot 1, 2 \cdot 3,</math> and <math>2 \cdot 5</math> are three valid choices for the <math>(2 \cdot odd)</math> factor above.
  
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The number of switches in position A is:
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:We have <math>{3\choose{1}} \cdot 3 \cdot 5^{2}= 225</math> ways.
  
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<math>1000-125-225 = 650</math>.
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The number of switches in position A is <math>1000-125-225 = \boxed{650}</math>.
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]
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{{MAA Notice}}

Latest revision as of 12:33, 4 July 2016

Problem

There is a set of 1000 switches, each of which has four positions, called $A, B, C$, and $D$. When the position of any switch changes, it is only from $A$ to $B$, from $B$ to $C$, from $C$ to $D$, or from $D$ to $A$. Initially each switch is in position $A$. The switches are labeled with the 1000 different integers $(2^{x})(3^{y})(5^{z})$, where $x, y$, and $z$ take on the values $0, 1, \ldots, 9$. At step i of a 1000-step process, the $i$-th switch is advanced one step, and so are all the other switches whose labels divide the label on the $i$-th switch. After step 1000 has been completed, how many switches will be in position $A$?

Solution

For each $i$th switch (designated by $x_{i},y_{i},z_{i}$), it advances itself only one time at the $i$th step; thereafter, only a switch with larger $x_{j},y_{j},z_{j}$ values will advance the $i$th switch by one step provided $d_{i}= 2^{x_{i}}3^{y_{i}}5^{z_{i}}$ divides $d_{j}= 2^{x_{j}}3^{y_{j}}5^{z_{j}}$. Let $N = 2^{9}3^{9}5^{9}$ be the max switch label. To find the divisor multiples in the range of $d_{i}$ to $N$, we consider the exponents of the number $\frac{N}{d_{i}}= 2^{9-x_{i}}3^{9-y_{i}}5^{9-z_{i}}$. In general, the divisor-count of $\frac{N}{d}$ must be a multiple of 4 to ensure that a switch is in position A:

$4n = [(9-x)+1] [(9-y)+1] [(9-z)+1] = (10-x)(10-y)(10-z)$, where $0 \le x,y,z \le 9.$

We consider the cases where the 3 factors above do not contribute multiples of 4.

  • Case of no 2's:
The switches must be $(\mathrm{odd})(\mathrm{odd})(\mathrm{odd})$. There are $5$ odd integers in $0$ to $9$, so we have $5 \times 5 \times 5 = 125$ ways.
  • Case of a single 2:
The switches must be one of $(2\cdot \mathrm{odd})(\mathrm{odd})(\mathrm{odd})$ or $(\mathrm{odd})(2 \cdot \mathrm{odd})(\mathrm{odd})$ or $(\mathrm{odd})(\mathrm{odd})(2 \cdot \mathrm{odd})$.
Since $0 \le x,y,z \le 9,$ the terms $2\cdot 1, 2 \cdot 3,$ and $2 \cdot 5$ are three valid choices for the $(2 \cdot odd)$ factor above.
We have ${3\choose{1}} \cdot 3 \cdot 5^{2}= 225$ ways.

The number of switches in position A is $1000-125-225 = \boxed{650}$.

See also

1999 AIME (Problems • Answer Key • Resources)
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Problem 8
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