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Difference between revisions of "2000 AMC 10 Problems/Problem 12"

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==Problem==
 
==Problem==
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Figures <math>0</math>, <math>1</math>, <math>2</math>, and <math>3</math> consist of <math>1</math>, <math>5</math>, <math>13</math>, and <math>25</math> nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
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<asy>
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unitsize(8);
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draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
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draw((9,0)--(10,0)--(10,3)--(9,3)--cycle);
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draw((8,1)--(11,1)--(11,2)--(8,2)--cycle);
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draw((19,0)--(20,0)--(20,5)--(19,5)--cycle);
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draw((18,1)--(21,1)--(21,4)--(18,4)--cycle);
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draw((17,2)--(22,2)--(22,3)--(17,3)--cycle);
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draw((32,0)--(33,0)--(33,7)--(32,7)--cycle);
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draw((29,3)--(36,3)--(36,4)--(29,4)--cycle);
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draw((31,1)--(34,1)--(34,6)--(31,6)--cycle);
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draw((30,2)--(35,2)--(35,5)--(30,5)--cycle);
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label("Figure",(0.5,-1),S);
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label("$0$",(0.5,-2.5),S);
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label("Figure",(9.5,-1),S);
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label("$1$",(9.5,-2.5),S);
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label("Figure",(19.5,-1),S);
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label("$2$",(19.5,-2.5),S);
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label("Figure",(32.5,-1),S);
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label("$3$",(32.5,-2.5),S);
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</asy>
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<math>\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801</math>
  
 
==Solution==
 
==Solution==

Revision as of 20:49, 9 January 2009

Problem

Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?

[asy] unitsize(8); draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((9,0)--(10,0)--(10,3)--(9,3)--cycle); draw((8,1)--(11,1)--(11,2)--(8,2)--cycle); draw((19,0)--(20,0)--(20,5)--(19,5)--cycle); draw((18,1)--(21,1)--(21,4)--(18,4)--cycle); draw((17,2)--(22,2)--(22,3)--(17,3)--cycle); draw((32,0)--(33,0)--(33,7)--(32,7)--cycle); draw((29,3)--(36,3)--(36,4)--(29,4)--cycle); draw((31,1)--(34,1)--(34,6)--(31,6)--cycle); draw((30,2)--(35,2)--(35,5)--(30,5)--cycle); label("Figure",(0.5,-1),S); label("$0$",(0.5,-2.5),S); label("Figure",(9.5,-1),S); label("$1$",(9.5,-2.5),S); label("Figure",(19.5,-1),S); label("$2$",(19.5,-2.5),S); label("Figure",(32.5,-1),S); label("$3$",(32.5,-2.5),S); [/asy]


$\mathrm{(A)}\ 10401 \qquad\mathrm{(B)}\ 19801 \qquad\mathrm{(C)}\ 20201 \qquad\mathrm{(D)}\ 39801 \qquad\mathrm{(E)}\ 40801$

Solution

We have a recursion:

$A_n=A_{n-1}+4(n-1)$.

I.E. we add increasing multiples of $4$ each time we go up a figure.

So, to go from Figure 0 to 100, we add

$4 \cdot 1+4 \cdot 2+...+4 \cdot 99=4 \cdot 4950=19800$.


$19801$

B.

See Also

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
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
All AMC 10 Problems and Solutions
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