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Difference between revisions of "2008 AMC 12B Problems/Problem 2"
m (Typos and some clarification.)
 
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\end{tabular}</math>
 
\end{tabular}</math>
  
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The positive difference between the two diagonal sums is then <math>(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{\textbf{(B) }}4</math>.
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The positive difference between the two diagonal sums is then <math>(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{\textbf{(B) } 4}</math>.
  
 
==Solution 2==
 
==Solution 2==
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You can also see that the positive difference between <math>9</math> and <math>10</math> is <math>1</math> a d the positive difference between <math>22</math> and <math>25</math> is <math>3</math>. Adding gives <math>1+3=\boxed{\textbf{(B) }}4</math>
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Notice that at baseline the diagonals sum to the same number (<math>52</math>).  Therefore we need only compute the effect of the swap.  The positive difference between <math>9</math> and <math>10</math> is <math>1</math> and the positive difference between <math>22</math> and <math>25</math> is <math>3</math>. Adding gives <math>1+3=\boxed{\textbf{(B) } 4}</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 22:53, 31 July 2023

The following problem is from both the 2008 AMC 12B #2 and 2008 AMC 10B #2, so both problems redirect to this page.

Problem

A $4\times 4$ block of calendar dates is shown. First, the order of the numbers in the second and the fourth rows are reversed. Then, the numbers on each diagonal are added. What will be the positive difference between the two diagonal sums?

$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{tabular}$

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 10$

Solution 1

After reversing the numbers on the second and fourth rows, the block will look like this:

$\begin{tabular}[t]{|c|c|c|c|} \multicolumn{4}{c}{}\\\hline 1&2&3&4\\\hline 11&10&9&8\\\hline 15&16&17&18\\\hline 25&24&23&22\\\hline \end{tabular}$

The positive difference between the two diagonal sums is then $(4+9+16+25)-(1+10+17+22)=3-1-1+3=\boxed{\textbf{(B) } 4}$.

Solution 2

Notice that at baseline the diagonals sum to the same number ($52$). Therefore we need only compute the effect of the swap. The positive difference between $9$ and $10$ is $1$ and the positive difference between $22$ and $25$ is $3$. Adding gives $1+3=\boxed{\textbf{(B) } 4}$

See Also

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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
2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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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