Difference between revisions of "1993 AIME Problems/Problem 9"
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'''Note:''' One can just substitute <math>1993\equiv-7\pmod{2000}</math> and <math>1994\equiv-6\pmod{2000}</math> to simplify calculations. | '''Note:''' One can just substitute <math>1993\equiv-7\pmod{2000}</math> and <math>1994\equiv-6\pmod{2000}</math> to simplify calculations. | ||
| − | + | == Solution 2 == | |
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Two labels <math>a</math> and <math>b</math> occur on the same point if <math>\ a(a+1)/2\equiv \ b(b+1)/2\ pmod{2000}</math>. If we assume the final answer be <math>n</math>, then we have <math>\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}</math>. | Two labels <math>a</math> and <math>b</math> occur on the same point if <math>\ a(a+1)/2\equiv \ b(b+1)/2\ pmod{2000}</math>. If we assume the final answer be <math>n</math>, then we have <math>\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}</math>. | ||
Revision as of 01:04, 22 September 2023
Contents
Problem
Two thousand points are given on a circle. Label one of the points 
. From this point, count 
points in the clockwise direction and label this point . From the point labeled , count 
points in the clockwise direction and label this point . (See figure.) Continue this process until the labels 
are all used. Some of the points on the circle will have more than one label and some points will not have a label. What is the smallest integer that labels the same point as 
?
Solution
The label will occur on the 
th point around the circle. (Starting from 1) A number 
will only occupy the same point on the circle if 
.
Simplifying this expression, we see that 
. Therefore, one of 
or 
is odd, and each of them must be a multiple of 
or 
.
For to be a multiple of and to be a multiple of , 
and 
. The smallest for this case is 
.
In order for to be a multiple of and to be a multiple of , 
and 
. The smallest for this case is larger than , so 
is our answer.
Note: One can just substitute 
and 
to simplify calculations.
Solution 2
Two labels 
and 
occur on the same point if 
. If we assume the final answer be , then we have .
Multiply on both side we have 
. As they have different parities, the even one must be divisible by 
.As 
, one of them is divisible by 
, which indicates it's divisible by .
Which leads to four different cases: 
; 
; 
and 
; 
and 
. Which leads to 
and 
respectively, and only 
satisfied.Therefore answer is .
See also
| 1993 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 8 |
Followed by Problem 10 | |
| 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. 
