Difference between revisions of "1999 AIME Problems/Problem 7"
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== Solution == | == 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: | ||
+ | |||
+ | <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. | ||
+ | |||
+ | *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 == | == See also == | ||
− | + | {{AIME box|year=1999|num-b=6|num-a=8}} | |
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 13:33, 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:
![$4n = [(9-x)+1] [(9-y)+1] [(9-z)+1] = (10-x)(10-y)(10-z)$](http://latex.artofproblemsolving.com/3/2/3/323a7b0ea0207ff505c100ebd4d9c498f2abdab0.png)
![$0 \le x,y,z \le 9.$](http://latex.artofproblemsolving.com/2/c/1/2c1ff361cb4b66d12f53ab5bff125b0bffeabb6f.png)
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.