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

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== See also ==
 
== See also ==
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* [[1999_AIME_Problems/Problem_6|Previous Problem]]
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* [[1999_AIME_Problems/Problem_8|Next Problem]]
 
* [[1999 AIME Problems]]
 
* [[1999 AIME Problems]]

Revision as of 00:51, 22 January 2007

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:

$(odd)(odd)(odd)$

There are $5$ odd integers in $0$ to $9$.

We have $5 \times 5 \times 5 = 5^{3}= 125$ ways.


Case of a single 2:

$(2\cdot odd)(odd)(odd)$ or $(odd)(2 \cdot odd)(odd)$ or $(odd)(odd)(2 \cdot odd)$

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:

$1000-125-225 = 650$.

See also

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