2000 AMC 12 Problems/Problem 8

The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page.

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 1

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 $n=3$ :

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

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 $\boxed{\textbf{(C) }20201}$ .

Solution 2

Using the recursion from solution 1, we see that the first differences of $4, 8, 12, ...$ form an arithmetic progression, and consequently that the second differences are constant and all equal to $4$ . Thus, the original sequence can be generated from a quadratic function.

If $f(n) = an^2 + bn + c$ , and $f(0) = 1$ , $f(1) = 5$ , and $f(2) = 13$ , we get a system of three equations in three variables:

$f(0) = 1$ gives $c = 1$

$f(1) = 5$ gives $a + b + c = 5$

$f(2) = 13$ gives $4a + 2b + c = 13$

Plugging in $c=1$ into the last two equations gives

$a + b = 4$

$4a + 2b = 12$

Dividing the second equation by 2 gives the system:

$a + b = 4$

$2a + b = 6$

Subtracting the first equation from the second gives $a = 2$ , and hence $b = 2$ . Thus, our quadratic function is:

$f(n) = 2n^2 + 2n + 1$

Calculating the answer to our problem, $f(100) = 20000 + 200 + 1 = 20201$ , which is choice $\boxed{\textbf{(C) }20201}$ .

Solution 3

We can see that each figure $n$ has a central box and 4 columns of $n$ boxes on each side of each square. Therefore, at figure 100, there is a central box with 100 boxes on the top, right, left, and bottom. Knowing that each quarter of each figure has a pyramid structure, we know that for each quarter there are $\sum_{n=1}^{100} n = 5050$ squares. $4 \cdot 5050 = 20200$ . Adding in the original center box we have $20200 + 1 = \boxed{\textbf{(C) }20201}$ .

Solution 5

Let $a_n$ be the number of squares in figure $n$ . We can easily see that $a_0=4\cdot 0+1$ $a_1=4\cdot 1+1$ $a_2=4\cdot 3+1$ $a_3=4\cdot 6+1.$ Note that in $a_n$ , the number multiplied by the 4 is the $n$ th triangular number. Hence, $a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}$ .

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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 12 Problems and Solutions

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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