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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 ==
 
== 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>.
+
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>.
+
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</math> and <math>118\pmod{4000}</math> respectively, and only <math>n=118</math> satisfied.Therefore answer is <math>\boxed{118}</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</math> and <math>118\pmod{4000}</math> respectively, and only <math>n=118</math> satisfied.Therefore answer is <math>\boxed{118}</math>.(by ZJY)
  
 
== See also ==
 
== See also ==

Latest revision as of 01:06, 22 September 2023

Problem

Two thousand points are given on a circle. Label one of the points $1$. From this point, count $2$ points in the clockwise direction and label this point $2$. From the point labeled $2$, count $3$ points in the clockwise direction and label this point $3$. (See figure.) Continue this process until the labels $1,2,3\dots,1993\,$ 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 $1993$?

AIME 1993 Problem 9.png

Solution

The label $1993$ will occur on the $\frac12(1993)(1994) \pmod{2000}$th point around the circle. (Starting from 1) A number $n$ will only occupy the same point on the circle if $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$.

Simplifying this expression, we see that $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{2000}$. Therefore, one of $1993 - n$ or $1994 + n$ is odd, and each of them must be a multiple of $125$ or $16$.

For $1993 - n$ to be a multiple of $125$ and $1994 + n$ to be a multiple of $16$, $n \equiv 118 \pmod {125}$ and $n\equiv 6 \pmod {16}$. The smallest $n$ for this case is $118$.

In order for $1993 - n$ to be a multiple of $16$ and $1994 + n$ to be a multiple of $125$, $n\equiv 9\pmod{16}$ and $n\equiv 6\pmod{125}$. The smallest $n$ for this case is larger than $118$, so $\boxed{118}$ is our answer.

Note: One can just substitute $1993\equiv-7\pmod{2000}$ and $1994\equiv-6\pmod{2000}$ to simplify calculations.

Solution 2

Two labels $a$ and $b$ occur on the same point if $\ a(a+1)/2\equiv \ b(b+1)/2\pmod{2000}$. If we assume the final answer be $n$, then we have $\frac12(n)(n + 1)\equiv \frac12(1993)(1994) \pmod{2000}$.

Multiply $2$ on both side we have $(1993)(1994) - (n)(n + 1) = (1993 - n)(1994 + n)\equiv 0\pmod{4000}$. As they have different parities, the even one must be divisible by $32$. As $(1993 - n)+(1994 + n)\equiv 2\pmod{5}$, one of them is divisible by $5$, which indicates it's divisible by $125$.

Which leads to four different cases: $1993-n\equiv 0\pmod{4000}$ ; $1994+n\equiv 0\pmod{4000}$ ; $1993-n\equiv 0\pmod{32}$ and $1994+n\equiv 0\pmod{125}$ ; $1993-n\equiv 0\pmod{125}$ and $1994+n\equiv 0\pmod{32}$. Which leads to $n\equiv 1993,2006,3881$ and $118\pmod{4000}$ respectively, and only $n=118$ satisfied.Therefore answer is $\boxed{118}$.(by ZJY)

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

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