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

Revision as of 22:22, 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

Solution 1

We have a recursion:

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

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

$4 \cdot 1+4 \cdot 2+...+4 \cdot 99+4\cdot 100=4 \cdot 5050=20200$ .

$20201$

$\boxed{\text{C}}$

Solution 2

We can divide up figure $n$ to get the sum of the sum of the first $n+1$ odd numbers and the sum of the first $n$ odd numbers. If you do not see this, here is the example for $n=3$ :

$[asy] draw((3,0)--(4,0)--(4,7)--(3,7)--cycle); draw((0,3)--(7,3)--(7,4)--(0,4)--cycle); draw((2,1)--(5,1)--(5,6)--(2,6)--cycle); draw((1,2)--(6,2)--(6,5)--(1,5)--cycle); draw((3,0)--(3,7),linewidth(1.5)); [/asy]$

The sum of the first $n$ odd numbers is $n^2$ , so for figure $n$ , there are $(n+1)^2+n^2$ unit squares. We plug in $n=100$ to get $20201$ , which is choice $\boxed{\text{C}}$

@@ Line 29: / Line 29: @@
 ==Solution==
+===Solution 1===
 We have a recursion:
-<math>A_n=A_{n-1}+4(n-1)</math>.
+<math>A_n=A_{n-1}+4n</math>.
 I.E. we add increasing multiples of <math>4</math> each time we go up a figure.
@@ Line 38: / Line 40: @@
 So, to go from Figure 0 to 100, we add
-<math>4 \cdot 1+4 \cdot 2+...+4 \cdot 99=4 \cdot 4950=19800</math>.
+<math>4 \cdot 1+4 \cdot 2+...+4 \cdot 99+4\cdot 100=4 \cdot 5050=20200</math>.
+<math>20201</math>
+<math>\boxed{\text{C}}</math>
+===Solution 2===
+We can divide up figure <math>n</math> to get the sum of the sum of the first <math>n+1</math> odd numbers and the sum of the first <math>n</math> odd numbers. If you do not see this, here is the example for <math>n=3</math>:
-<math>19801</math>
+<asy>
+draw((3,0)--(4,0)--(4,7)--(3,7)--cycle);
+draw((0,3)--(7,3)--(7,4)--(0,4)--cycle);
+draw((2,1)--(5,1)--(5,6)--(2,6)--cycle);
+draw((1,2)--(6,2)--(6,5)--(1,5)--cycle);
+draw((3,0)--(3,7),linewidth(1.5));
+</asy>
-B.
+The sum of the first <math>n</math> odd numbers is <math>n^2</math>, so for figure <math>n</math>, there are <math>(n+1)^2+n^2</math> unit squares. We plug in <math>n=100</math> to get <math>20201</math>, which is choice <math>\boxed{\text{C}}</math>
 ==See Also==
 {{AMC10 box|year=2000|num-b=11|num-a=13}}

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