Difference between revisions of "2000 AMC 12 Problems/Problem 8"

Revision as of 04:48, 3 November 2021

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]$

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

Video:

https://www.youtube.com/watch?v=HVP6qjKAkjA&t=2s

.

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 4

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, $a_{100}=4\cdot \frac{100\cdot 101}{2}+1=\boxed{\textbf{(C) }20201}$ . ~qkddud~

Solution 5

Let $f_n$ denote the number of unit cubes in a figure. We have $f_0=1$ $f_1=5$ $f_2=13$ $f_3=25$ $f_4=41$ $...$

Computing the difference between the number of cubes in each figure yields $4,8,12,16,...$ It is easy to notice that this is an arithmetic sequence, with the first term being $4$ and the difference being $4$ . Let this sequence be $a_n$

From $f_0$ to $f_{100}$ , the sequence will have $100$ terms. Using the arithmetic sum formula yields

$S_{100}=\frac{100[2\cdot 4+(100-1)4]}{2}$ $=50(2\cdot 4+99\cdot 4)$ $=50(101\cdot 4)$ $=200\cdot 101$ $=20200$

So $f_{100}=1+20200=\boxed{\textbf{(C) }20201}$ unit cubes.

~ljlbox

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 70: / Line 70: @@
 ~ljlbox
-== Solution 6(Newton's Forward Differences) ==
-We know that 1 and 5 differ by 4, 5 and 13 differ by 8, and 13 and 25 differ by 12. Hence the differences are 4, 8, and 12, resp. And the differences of the differences area all 4. So by Newton's Forward Difference Formula, we get the 100th figure is(because Figure 0 exists) <math>\dbinom{101-1}{0}+4\dbinom{101-1}{1}+4\dbinom{101-1}{2}=20201</math> or <math>\textbf{(C)}</math>
--vsamc
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