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 == | ||
+ | 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>. | ||
+ | |||
+ | Multiply <math>2</math> on both side we have <math>(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{4000}</math>. As they have different parities, the even one must be divisible by <math>32</math>.As <math> (1993 - n)+(1994 + n)\equiv 2\pmod{5}</math>, one of them is divisible by <math>5</math>, which indicates it's divisible by <math>125</math>. | ||
+ | |||
+ | Which leads to four different cases: <math>1993-n\equiv 0\pmod{4000}</math> ; <math>1994+n\equiv 0\pmod{4000}</math> ; <math>1993-n\equiv 0\pmod{32}</math> and <math>1994+n\equiv 0\pmod{125}</math> ; <math>1993-n\equiv 0\pmod{125}</math> and <math>1994+n\equiv 0\pmod{32}</math>. Which leads to <math>n\equiv 1993,2006,3881 and 118\pmod{4000}</math> respectively, and only <math>n=118</math> satisfied.Therefore answer is <math>\boxed{118}</math>. | ||
== See also == | == See also == |
Revision as of 02:03, 22 September 2023
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
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.