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

Revision as of 16:40, 12 August 2006

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:

$1000-125-225 = 650$ .

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 == Solution ==
+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:
+<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>
+We consider the cases where the 3 factors above do not contribute multiples of 4.
+Case of no 2's:
+<math>(odd)(odd)(odd)</math>
+There are <math>5</math> odd integers in <math>0</math> to <math>9</math>.
+We have <math>5 \times 5 \times 5 = 5^{3}= 125</math> ways.
+Case of a single 2:
+<math>(2\cdot odd)(odd)(odd)</math> or <math>(odd)(2 \cdot odd)(odd)</math> or <math>(odd)(odd)(2 \cdot odd)</math>
+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.
+We have <math>{3\choose{1}} \cdot 3 \cdot 5^{2}= 225</math> ways.
+The number of switches in position A is:
+<math>1000-125-225 = 650</math>.
 == See also ==
 * [[1999 AIME Problems]]

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

Revision as of 16:40, 12 August 2006

Problem

Solution

See also