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: | ||
− | <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><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: | ||
− | + | :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. | |
− | + | *Case of a single 2: | |
− | + | :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>. | |
− | + | :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 = \boxed{650}</math>. | ||
− | + | == See also == | |
− | + | {{AIME box|year=1999|num-b=6|num-a=8}} | |
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− | + | [[Category:Intermediate Combinatorics Problems]] | |
− | + | {{MAA Notice}} | |
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Latest revision as of 12:33, 4 July 2016
Problem
There is a set of 1000 switches, each of which has four positions, called , and . When the position of any switch changes, it is only from to , from to , from to , or from to . Initially each switch is in position . The switches are labeled with the 1000 different integers , where , and take on the values . At step i of a 1000-step process, the -th switch is advanced one step, and so are all the other switches whose labels divide the label on the -th switch. After step 1000 has been completed, how many switches will be in position ?
Solution
For each th switch (designated by ), it advances itself only one time at the th step; thereafter, only a switch with larger values will advance the th switch by one step provided divides . Let be the max switch label. To find the divisor multiples in the range of to , we consider the exponents of the number . In general, the divisor-count of must be a multiple of 4 to ensure that a switch is in position A:
We consider the cases where the 3 factors above do not contribute multiples of 4.
- Case of no 2's:
- The switches must be . There are odd integers in to , so we have ways.
- Case of a single 2:
- The switches must be one of or or .
- Since the terms and are three valid choices for the factor above.
- We have ways.
The number of switches in position A is .
See also
1999 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.