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Difference between revisions of "1989 AHSME Problems/Problem 23"
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so the highest value is <math>n=43</math> for which the sum is <math>1935</math>.
 
so the highest value is <math>n=43</math> for which the sum is <math>1935</math>.
  
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After <math>1935</math> minutes the 43rd square is enclosed and the particle is at the point <math>(0,43)</math>. During the 1936th minute it moves up to <math>(0,44)</math>. At the end of the 1980th minute it has moved right to <math>(44,44)</math>. After this it moves downward, and at the end of the 1989th minute it is at <math>(44,35)</math>.
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After <math>1935</math> minutes the 43rd square is enclosed and the particle is at the point <math>(0,43)</math>. During the 1936th minute it moves up to <math>(0,44)</math>. At the end of the 1980th minute it has moved right to <math>(44,44)</math>. After this it moves downward, and at the end of the 1989th minute it is at <math>(44,35)</math>. The answer is <math>\boxed{\textbf{(D)}}</math>.
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== See also ==
 +
{{AHSME box|year=1989|num-b=22|num-a=24}} 
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[[Category: Intermediate Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 13:24, 21 June 2023

Problem

A particle moves through the first quadrant as follows. During the first minute it moves from the origin to $(1,0)$. Thereafter, it continues to follow the directions indicated in the figure, going back and forth between the positive x and y axes, moving one unit of distance parallel to an axis in each minute. At which point will the particle be after exactly 1989 minutes?

[asy] import graph; Label f; f.p=fontsize(6); xaxis(0,3.5,Ticks(f, 1.0)); yaxis(0,4.5,Ticks(f, 1.0)); draw((0,0)--(1,0)--(1,1)--(0,1)--(0,2)--(2,2)--(2,0)--(3,0)--(3,3)--(0,3)--(0,4)--(1.5,4),blue+linewidth(2)); arrow((2,4),dir(180),blue); [/asy]

$\text{(A)}\ (35,44)\qquad\text{(B)}\ (36,45)\qquad\text{(C)}\ (37,45)\qquad\text{(D)}\ (44,35)\qquad\text{(E)}\ (45,36)$

Solution

Squares of size $1\times1,\ 2\times2,\ 3\times3,\ ...$ are successively enclosed between the path and the axes.

[asy] import graph; axialshade((0,0)--(0,3)--(3,3)--(3,0)--cycle,yellow,(-6,-6),white,(3,3)); axialshade((0,0)--(0,2)--(2,2)--(2,0)--cycle,orange,(-3,-3),white,(2,2)); axialshade((0,0)--(0,1)--(1,1)--(1,0)--cycle,red,(-2,-2),white,(1,1)); Label f; f.p=fontsize(6); xaxis(0,3.5,Ticks(f, 1.0)); yaxis(0,4.5,Ticks(f, 1.0)); draw((0,0)--(1,0)--(1,1)--(0,1),red+linewidth(2)); draw((0,1)--(0,2)--(2,2)--(2,0),orange+linewidth(2)); draw((2,0)--(3,0)--(3,3)--(0,3),yellow+linewidth(2)); draw((0,3)--(0,4)--(1.5,4),green+linewidth(2)); arrow((2,4),dir(180),green); dot((0,1),red);dot((2,0),orange);dot((0,3),yellow); [/asy]

It takes $1+1+1$ minutes to enclose the first square, $1+2+2$ minutes to enclose the second, $1+3+3$ minutes to enclose the third, and so on. After odd squares, the particle is on the Y axis; after even squares, the particle is on the X axis.

First we find the highest integer $n$ such that $\sum_{k=1}^n1+2k\le1989$. The sum is equal to \[\sum_{k=1}^n1+2\sum_{k=1}^nk=n+n(n+1)=n(n+2)=(n+1)^2-1\] so the highest value is $n=43$ for which the sum is $1935$.

After $1935$ minutes the 43rd square is enclosed and the particle is at the point $(0,43)$. During the 1936th minute it moves up to $(0,44)$. At the end of the 1980th minute it has moved right to $(44,44)$. After this it moves downward, and at the end of the 1989th minute it is at $(44,35)$. The answer is $\boxed{\textbf{(D)}}$.

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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