AMC 10 Step by Step

2008 AMC 10B Problem 2

P1-10~2 minPrint

A 4×44\times 4 block of calendar dates is shown. The order of the numbers in the second row is to be reversed. Then the order of the numbers in the fourth row is to be reversed. Finally, the numbers on each diagonal are to be added. What will be the positive difference between the two diagonal sums?

12348910111516171822232425\begin{array}{|c|c|c|c|} \hline 1&2&3&4\\\hline 8&9&10&11\\\hline 15&16&17&18\\\hline 22&23&24&25\\\hline \end{array}

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Key insight

Write out the new block after reversing the two rows, then add each diagonal. The sums are 50 and 54.

Solution

This is a bookkeeping problem, so the safest plan is to actually write the new block down instead of keeping it in your head.

1. Reverse the second row: 8,9,10,118, 9, 10, 11 becomes 11,10,9,811, 10, 9, 8.
2. Reverse the fourth row: 22,23,24,2522, 23, 24, 25 becomes 25,24,23,2225, 24, 23, 22.
3. The new block is
12341110981516171825242322\begin{array}{cccc} 1 & 2 & 3 & 4 \\ 11 & 10 & 9 & 8 \\ 15 & 16 & 17 & 18 \\ 25 & 24 & 23 & 22 \end{array}
4. Diagonal from top-left to bottom-right: 1+10+17+22=501 + 10 + 17 + 22 = 50.
5. Diagonal from top-right to bottom-left: 4+9+16+25=544 + 9 + 16 + 25 = 54.
6. The difference is 54−50=454 - 50 = 4.

The answer is (B) 4\boxed{\textbf{(B)}\ 4}.

Why this works

Nothing clever is needed; the only danger is a copying slip, and writing the new block down removes it. You can double-check with a shortcut. In the original block both diagonals add to 5252 (1+9+17+251 + 9 + 17 + 25 and 4+10+16+224 + 10 + 16 + 22). After the reversals, the top-left diagonal gets 1010 instead of 99 (up 11) and 2222 instead of 2525 (down 33), so it drops by 22 to 5050. The other diagonal rises by 22 to 5454.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Reverse the second row (8,9,10,11→11,10,9,88, 9, 10, 11 \to 11, 10, 9, 8) and the fourth row (22,23,24,25→25,24,23,2222, 23, 24, 25 \to 25, 24, 23, 22). The block becomes
12341110981516171825242322\begin{array}{cccc} 1 & 2 & 3 & 4 \\ 11 & 10 & 9 & 8 \\ 15 & 16 & 17 & 18 \\ 25 & 24 & 23 & 22 \end{array}

Main diagonal (top-left to bottom-right): 1+10+17+22=501 + 10 + 17 + 22 = 50.

Other diagonal (top-right to bottom-left): 4+9+16+25=544 + 9 + 16 + 25 = 54.

The positive difference is 54−50=454 - 50 = 4.

The answer is (B) 4\boxed{\textbf{(B)}\ 4}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.