1999 AIME Problems/Problem 7

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 $1000-125-225 = \boxed{650}$ .

1999 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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1999 AIME Problems/Problem 7

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