Difference between revisions of "2002 AMC 8 Problems/Problem 11"

Latest revision as of 13:30, 29 October 2024

Problem

A sequence of squares is made of identical square tiles. The edge of each square is one tile length longer than the edge of the previous square. The first three squares are shown. How many more tiles does the seventh square require than the sixth?

$[asy] path p=origin--(1,0)--(1,1)--(0,1)--cycle; draw(p); draw(shift(3,0)*p); draw(shift(3,1)*p); draw(shift(4,0)*p); draw(shift(4,1)*p); draw(shift(7,0)*p); draw(shift(7,1)*p); draw(shift(7,2)*p); draw(shift(8,0)*p); draw(shift(8,1)*p); draw(shift(8,2)*p); draw(shift(9,0)*p); draw(shift(9,1)*p); draw(shift(9,2)*p);[/asy]$

$\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15$

Solutions

Solution 1

The first square has a sidelength of $1$ , the second square $2$ , and so on. The seventh square has $7$ and is made of $7^2=49$ unit tiles. The sixth square has $6$ and is made of $6^2=36$ unit tiles. The seventh square has $49-36=\boxed{\text{(C)}\ 13}$ more tiles than the sixth square.

Solution 2

The edge of each square is one tile length longer than the edge of the previous square, which means that each square has $2*$ edge length $- 1$ more tiles than the previous square, because each square is just one edge added on the top and on the right to the previous square, with one overlapping tile. Then the seventh square has $2*7-1=13$ more tiles than the sixth square, which is $\boxed{\text{(C)}\ 13}$ .

Solution 3

We see a pattern of 1, 4, and 9, all of which are the squares of 1, 2,and 3 respectively. So, the 6th and 7th squares will also follow the same pattern. Via the difference of squares, we see that $7^2 - 6^2$ . Now we can see that the seventh square has $(7-6)(7+6) =$ $\boxed{\text{(C)}\ 13}$ more tiles than the sixth square.

~ AkCANdo

Video Solution by Whymath

https://youtu.be/QGxNCcUB5zw

@@ Line 22: / Line 22: @@
 <math>\text{(A)}\ 11 \qquad \text{(B)}\ 12 \qquad \text{(C)}\ 13 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 15</math>
-==Solution==
+==Solutions==
+===Solution 1===
 The first square has a sidelength of <math>1</math>, the second square <math>2</math>, and so on. The seventh square has <math>7</math> and is made of <math>7^2=49</math> unit tiles. The sixth square has <math>6</math> and is made of <math>6^2=36</math> unit tiles. The seventh square has <math>49-36=\boxed{\text{(C)}\ 13}</math> more tiles than the sixth square.
+===Solution 2===
+The edge of each square is one tile length longer than the edge of the previous square, which means that each square has <math>2*</math> edge length <math>- 1</math> more tiles than the previous square, because each square is just one edge added on the top and on the right to the previous square, with one overlapping tile. Then the seventh square has <math>2*7-1=13</math> more tiles than the sixth square, which is <math>\boxed{\text{(C)}\ 13}</math>.
+===Solution 3===
+We see a pattern of 1, 4, and 9, all of which are the squares of 1, 2,and 3 respectively. So, the 6th and 7th squares will also follow the same pattern. Via the difference of squares, we see that <math>7^2 - 6^2</math>. Now we can see that the seventh square has  <math>(7-6)(7+6) = </math> <math>\boxed{\text{(C)}\ 13}</math> more tiles than the sixth square.
+~ AkCANdo
+==Video Solution by Whymath==
+https://youtu.be/QGxNCcUB5zw
 ==See Also==
 {{AMC8 box|year=2002|num-b=10|num-a=12}}
 {{MAA Notice}}

2002 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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 AJHSME/AMC 8 Problems and Solutions

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