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2016 AMC 12B Problems/Problem 12: Difference between revisions

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==Problem==
==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 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?
All the numbers <imath>1, 2, 3, 4, 5, 6, 7, 8, 9</imath> are written in a <imath>3\times3</imath> 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 <imath>18</imath>. 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>
<imath>\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9</imath>


==Solution==
==Solution==


Solution by Mlux:
Solution by Mlux:
Draw a <math>3\times3</math> 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 <math>18</math>. Trying <math>1+3+5+9 = 18</math> works. Place each odd number in the corner in a clockwise order. Then fill in the spaces. There has to be a <math>2</math> in between the <math>1</math> and <math>3</math>. There is a <math>4</math> between <math>3</math> and <math>5</math>.  The final grid should similar to this.
Draw a <imath>3\times3</imath> 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 <imath>18</imath>. Trying <imath>1+3+5+9 = 18</imath> works. Place each odd number in the corner in a clockwise order. Then fill in the spaces. There has to be a <imath>2</imath> in between the <imath>1</imath> and <imath>3</imath>. There is a <imath>4</imath> between <imath>3</imath> and <imath>5</imath>.  The final grid should similar to this.


<math>\newline
<imath>\newline
1, 2,  3\newline
1, 2,  3\newline
8,  7, 4\newline
8,  7, 4\newline
9,  6,  5</math>
9,  6,  5</imath>


<math>\textbf{(C)}7</math> is in the middle.
<imath>\textbf{(C)}7</imath> is in the middle.
==Solution 2==
==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 <math>25 - 18 = 7</math>.
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 <imath>25 - 18 = 7</imath>.
 
 
==Solution 3==
Note that if a number other than <imath>1</imath> or <imath>9</imath> is in a corner, the two numbers adjacent to it must be the numbers before and after it. Thus, if <imath>1</imath> is in an edge spot that isn't a corner, it will be adjacent to two corners, which isn't possible, because <imath>1</imath> is not consecutive to two other numbers between <imath>1</imath> and <imath>9</imath>. Similarly, <imath>9</imath> can't go to an edge spot that isn't a corner. Therefore, two of the corner squares are <imath>1</imath> and <imath>9</imath>.
 
Filling in the table one by one, we can start with this WLOG:
 
<imath>\newline
1, 2, \_ \newline
\_,  \_, \_ \newline
\_,  \_, \_ </imath>
 
Since <imath>3</imath> is not one of the answer choices, it cannot be in the center, so it must go in the corner adjacent to <imath>2</imath>. That means the last corner is <imath>18 - 1 - 3 - 9 = 5</imath>. Thus, we can keep filling in the table to get:
 
<imath>\newline
1, 2,  3\newline
8,  7, 4\newline
9,  6,  5</imath>
 
And our answer is <imath>\textbf{(C)}7</imath>.
 
~~[[User:StrikingTsunamiYT|StrikingTsunami]]


==Video Solution by CanadaMath (Problem 11-20)==
==Video Solution by CanadaMath (Problem 11-20)==

Latest revision as of 00:15, 7 November 2025

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 $25 - 18 = 7$.


Solution 3

Note that if a number other than $1$ or $9$ is in a corner, the two numbers adjacent to it must be the numbers before and after it. Thus, if $1$ is in an edge spot that isn't a corner, it will be adjacent to two corners, which isn't possible, because $1$ is not consecutive to two other numbers between $1$ and $9$. Similarly, $9$ can't go to an edge spot that isn't a corner. Therefore, two of the corner squares are $1$ and $9$.

Filling in the table one by one, we can start with this WLOG:

$\newline 1, 2, \_ \newline \_, \_, \_ \newline \_, \_, \_$

Since $3$ is not one of the answer choices, it cannot be in the center, so it must go in the corner adjacent to $2$. That means the last corner is $18 - 1 - 3 - 9 = 5$. Thus, we can keep filling in the table to get:

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

And our answer is $\textbf{(C)}7$.

~~StrikingTsunami

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

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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