Difference between revisions of "1999 AIME Problems/Problem 7"
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*Case of a single 2: | *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})(2 \cdot \mathrm{odd})</math>. | + | :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. | :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. | ||
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:We have <math>{3\choose{1}} \cdot 3 \cdot 5^{2}= 225</math> ways. | :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>. | + | The number of switches in position A is <math>1000-125-225 = \boxed{650}</math>. |
== See also == | == See also == | ||
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[[Category:Intermediate Combinatorics Problems]] | [[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.