2008 AMC 12B Problem 2

Attempt Problem 2 of the 2008 AMC 12B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 12B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

2.

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?
22
44
66
88
1010
Answer: B
Concepts:arithmetic sequencepattern recognition
Small hint:
After reversing rows 22 and 4,4, the rows become 1 2 3 4,1\,2\,3\,4, 11 10 9 8,11\,10\,9\,8, 15 16 17 18,15\,16\,17\,18, 25 24 23 2225\,24\,23\,22
Big hint:
One diagonal reads 1,10,17,221, 10, 17, 22 and the other reads 4,9,16,254, 9, 16, 25
Solution:
Reversing the second and fourth rows gives the array with rows 1 2 3 4,1\,2\,3\,4, 11 10 9 8,11\,10\,9\,8, 15 16 17 18,15\,16\,17\,18, and 25 24 23 22.25\,24\,23\,22.
The main diagonal sums to 1+10+17+22=50,1 + 10 + 17 + 22 = 50, and the other diagonal sums to 4+9+16+25=54.4 + 9 + 16 + 25 = 54.
The positive difference is 54−50=4.54 - 50 = 4.
Thus, the correct answer is B.
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