Difference between revisions of "2016 AMC 12B Problems/Problem 12"

Latest revision as of 00:02, 10 November 2024

Problem

All the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9$ are written in a $3\times3$ array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to $18$ . What is the number in the center?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$

Solution

Solution by Mlux: Draw a $3\times3$ matrix. Notice that no adjacent numbers could be in the corners since two consecutive numbers must share an edge. Now find 4 nonconsecutive numbers that add up to $18$ . Trying $1+3+5+9 = 18$ works. Place each odd number in the corner in a clockwise order. Then fill in the spaces. There has to be a $2$ in between the $1$ and $3$ . There is a $4$ between $3$ and $5$ . The final grid should similar to this.

$\newline 1, 2, 3\newline 8, 7, 4\newline 9, 6, 5$

$\textbf{(C)}7$ is in the middle.

Solution 2

If we color the square like a chessboard, since the numbers alternate between even and odd, and there are five odd numbers and four even numbers, the odd numbers must be in the corners/center, while the even numbers must be on the edges. Since the odd numbers add up to 25, and the numbers in the corners add up to 18, the number in the center must be $25 - 18 = 7$ .

Video Solution by CanadaMath (Problem 11-20)

Fast Forward to 6:20 for problem 12 https://www.youtube.com/watch?v=4osvFatUv1o

~THEMATHCANADIAN

@@ Line 1: / Line 1: @@
 ==Problem==
-All the numbers <math>1, 2, 3, 4, 5, 6, 7, 8, 9</math> are written in a <math>3\times3</math> array of squares, one number in each square, in such a way that if two numbers of consecutive then they occupy squares that share an edge. The numbers in the four corners add up to <math>18</math>. What is the number in the center?
+All the numbers <math>1, 2, 3, 4, 5, 6, 7, 8, 9</math> are written in a <math>3\times3</math> array of squares, one number in each square, in such a way that if two numbers are consecutive then they occupy squares that share an edge. The numbers in the four corners add up to <math>18</math>. What is the number in the center?
 <math>\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9</math>
@@ Line 17: / Line 17: @@
 <math>\textbf{(C)}7</math> is in the middle.
 ==Solution 2==
-If we color the square like a chessboard, since the numbers altrenate between even and odd, and there are five odd numbers and four even numbers, the odd numbers must be in the corners/center, while the even numbers must be on the edges.  Since the odd numbers add up to 25, and the numbers in the corners add up to 18, the number in the center must be 25-18=7
+If we color the square like a chessboard, since the numbers alternate between even and odd, and there are five odd numbers and four even numbers, the odd numbers must be in the corners/center, while the even numbers must be on the edges.  Since the odd numbers add up to 25, and the numbers in the corners add up to 18, the number in the center must be <math>25 - 18 = 7</math>.
+==Video Solution by CanadaMath (Problem 11-20)==
+Fast Forward to 6:20 for problem 12
+https://www.youtube.com/watch?v=4osvFatUv1o
+~THEMATHCANADIAN
 ==See Also==
 {{AMC12 box|year=2016|ab=B|num-b=11|num-a=13}}

2016 AMC 12B (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 12 Problems and Solutions

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