Difference between revisions of "2016 AMC 10B Problems/Problem 15"

(Solution 1 - Trial Error)
(Solution 2)
Line 17: Line 17:
 
<cmath>2  9  6</cmath>
 
<cmath>2  9  6</cmath>
 
<cmath>3  4  5</cmath>
 
<cmath>3  4  5</cmath>
with the numbers <math>1-8</math> around the outsides and <math>9</math> in the middle. We see that the sum of the four corner numbers is <math>16</math>. If we switch <math>7</math> and <math>9</math>, then the corner numbers will add up to <math>18</math> and the consecutive numbers will still be touching each other.
+
with the numbers <math>1-8</math> around the outsides and <math>9</math> in the middle. We see that the sum of the four corner numbers is <math>16</math>. If we switch <math>7</math> and <math>9</math>, then the corner numbers will add up to <math>18</math> and the consecutive numbers will still be touching each other. The answer is <math>\boxed{7}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2016|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2016|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 20:46, 21 February 2016

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 of 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 1 - Trial Error

Quick testing shows that \[3 2 1\] \[4 7 8\] \[5 6 9\] is a valid solution. $3+1+5+9 = 18$, and the numbers follow the given condition. The center number is found to be $\boxed{7}$. — @adihaya (talk) 12:27, 21 February 2016 (EST)

Solution 2

First let the numbers be \[1   8   7\] \[2   9   6\] \[3   4   5\] with the numbers $1-8$ around the outsides and $9$ in the middle. We see that the sum of the four corner numbers is $16$. If we switch $7$ and $9$, then the corner numbers will add up to $18$ and the consecutive numbers will still be touching each other. The answer is $\boxed{7}$.

See Also

2016 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
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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